Algebra

Algebra is the foundation of calculus. The basicidea behind algebra is rewriting equations and sim-plifying expressions; this includes such things as fac-toring, FOILing (i.e., (a+b)(c+d) = ac+ad+bc+bd),adding fractions (remember to get a common de-

nominator,a

b+c

d=ad+ bc

bd) and multiplying by

the conjugate (the conjugate of a + b is a − b andmultiplying gives a2 − b2). Below are some basicfacts that frequently come up.

(a+ b)(a− b) = a2 − b2

(a+ b)2 = a2 + 2ab+ b2

(a− b)2 = a2 − 2ab+ b2

Another useful fact is the quadratic formula (alsoknown as the quadratic equation)

ax2 + bx+ c = 0 ⇒ x =−b±

√b2 − 4ac

2a.

Rules for exponents and logarithms are useful:

xaxb = xa+b,xa

xb= xa−b, (xa)b = xab,

ln(ab) = lna+ lnb, ln(ab) = lna− lnb,

and ln(ab) = b lna.

Of course it is also useful to remember what wecannot do. For example, “

√x2 + a2 = x + a” is not

true, even when taking a test!Often associated with algebra are functions and

graphing. A function is a rule which takes an in-put and associates a unique output, i.e., y = f(x).The domain of a function are numbers that we canput into the function and get something out, whilethe range are the possible values that we can getout. Limitations on the domain arise from dividingby zero, taking square roots of negative numbers, orlog of a nonpositive number. Determining the rangeis hard, but fortunately there is calculus (woohoo!).Functions can be combined through addition, multi-plication, division, composition, etc.

We work with points in the plane by describingthem relative to how far away from the origin wehave moved in different directions, i.e., (x,y). Thedistance between the two points (x0,y0) and (x1,y1)in the plane is

distance =√

(x1 − x0)2 + (y1 − y0)2.

We have lines (y = mx + b where m is the slope(found by rise/run) and b the y-intercept), circles((x − x0)2 + (y − y0)

2 = r2 where (x0,y0) is the cen-ter of the circle and r the radius), parabolas (y =

ax2+bx+c), ellipses (ax2+by2 = 1) and hyperbolas(ax2 − by2 = 1).

The area of a rectangle is (length)×(width), thearea of a circle is πr2 (where r is the radius) and thearea of a triangle is 1

2(base)×(height). The volume of

a box is (length)×(width)×(height) and the volumeof a sphere is 4

3πr3 (again r is the radius).

Trigonometry

Trigonometry is based off the two ideas that (i) tri-angles are rigid and (ii) when we scale a triangle theratios of the sides do not change. Using these we canassociate values with the angles of triangles whichcan be used to solve various problems related to tri-angles. The basis of many important facts about tri-angles comes from the Pythagorean Theorem whichsays that for a right triangle with sides a,b, c (c beingthe hypotenuse) then a2 + b2 = c2. What this meansis that often when dealing with triangles we look forright triangles. Let θ be an angle of a right triangle,“opp” the length of the side opposite θ, “adj” thelength of the side adjacent to θ and “hyp” the lengthof the hypotenuse; then we have

sin θ =opphyp

, cos θ =adjhyp

,

tan θ =oppadj

, sec θ =hypadj

.

(We either measure angles in degrees (360◦ is a fullrevolution) or radians (2π is a full revolution). Incalculus we almost always will use radians.) Notethat the functions sin and cos are periodic and alsoare always bounded between −1 and 1 (← a veryuseful fact).

From the definitions have

tan θ =sin θcos θ

and sec θ =1

cos θ.

And by using the Pythagorean Theorem we can con-clude

sin2 θ+ cos2 θ = 1 and tan2 θ+ 1 = sec2 θ.

Other rules which frequently come up are the dou-ble angle formulas

sin(2θ) = 2 sin θ cos θcos(2θ) = cos2 θ− sin2 θ,

this can be used to reduce powers of sine and cosine

cos2 θ =1+ cos 2θ

2and sin2 θ =

1− cos 2θ2

We also have the even/odd’er identities

sin(−θ) = − sin θ, and cos(−θ) = cos θ.

There are of course many more formulas and iden-tities that arise in trigonometry, but for our purposesthese will usually suffice (we will make use, for ex-ample, of the law of cosines but in general these arethe most important identities).

Quiz 1 problem bank

1. Find the domain for the function

f(x) =

√9− x2

ln(2− ex).

2. Find f−1(x) given f(x) = 12(ex − e−x).

3. Find the slope and y-intercept for the liney = 5+ ln(7e5x).

4. Given f(x) = x+ 3 and g(x) = x2 + 4, thenh(x) = g(f(x)) − f(g(x)) is a line. Find the slopeand y-intercept for h(x).

5. Sketch the curve y = 2− x− |x| for −2 6 x 6 2.

6. Find all solutions (x,y) to the following:

xy− x+ 2y = 2

x2 + 5x+ 4y = 0

7. For −1 6 x 6 1, rewrite cos(2 arccos

√x+ 1

2

)as an algebraic expression.

8. For x > 1, rewrite cos(

arctan(

2x

x2 − 1

))as an

algebraic expression.

9. Given that the circles in the following diagramhave the same center and that the length of theline segment is 20, determine the area betweenthe circles.

10. Given for the diagram below thatAB = AC = CD and AD = BD, determine theangle α, give the answer in radians.

B

A

D

Cα

Rate of changeWe are interested in finding the rate of change of a

function. In particular, given a function y = f(x) weare interested in finding how fast y is changing withrespect to x at some fixed time x = a. The main waywe will do this is to combine two observations.

1. Given a line y = mx + b the slope is m and isfound by

m =riserun

=∆y

∆x=y2 − y1x2 − x1

.

In other words, for a line the rate of change of ywith respect to x is the slope.

2. For a typical function we will encounter, whenwe look at the function “near” x = a it will looklike a line, namely the tangent line. (The tangentline is the line which “touches without crossing”the curve.)

So to find the rate of change we will find the slope ofthe tangent line. But, to calculate the slope we needtwo points and the tangent line only gives one. Sofirst we work with the simpler case of the secant line(a secant line crosses the curve at two values of x).

The secant line which intersects the curve y = f(x)at

(a, f(a)

)and

(b, f(b)

)has slope

f(b) − f(a)

b− a,

or if we let b = a+ h this can be written as

f(a+ h) − f(a)

h.

This slope gives the average rate of change of y = f(x)from x = a to x = b, i.e., the constant rate thatf would have to change at to go from

(a, f(a)

)to(

b, f(b)).

The slope of the secant line will approximate theslope of the tangent line and the approximation willget better and better as b gets closer to a (or equiv-alently as h goes to 0). The problem is that if a = bor if h = 0 then these slopes are 0/0 which are unde-fined, so we need some way to handle this.

LimitsThe way we handle this is to use limits. Intuitively

limits tell us what should happen based on what ishappening nearby. So for example

limx→c

g(x) = L,

which we read “the limit as x goes to c of g(x) is L”,means that as x gets close to c the function g(x) isgetting close to L (and staying close!). It is possiblethat the limit does not exist. For example,

limx→0

sin(1

x) = Does not exist.

To see this we note that the function sin(1/x) willdo “infinitely” many oscillations between 1 and −1around x = 0 and so it does not approach a singlefixed L.

One way to guess a limit is to plug in values ofx closer and closer to c and see if it is approachingsome certain value; we can also try plotting a pictureof g(x) near x = c and seeing how the function isbehaving. Both of these methods have short comings,in particular they are hard to do without a calculatorand can sometimes be deceiving. So we want to havesome methods to deal with these limits. One methodis to build up a collection of rules that we can use.For example we have the following two rules

limx→c

k = k and limx→c

x = c.

(The first follows by noting that k is always close tok, and the second says “as x gets close to c then xgets close to c.) On the other hand we have that if

limx→c

f(x) = L and limx→c

g(x) =M

and L,M are finite then we have the following rules,which essentially say that limits do what we thinkthey should do.

1. limx→c

(f(x) + g(x)

)=

(limx→c

f(x))+(

limx→c

g(x)).

2. limx→c

(kf(x)

)= k

(limx→c

f(x))

3. limx→c

(f(x)g(x)

)=

(limx→c

f(x))(

limx→c

g(x)).

4. limx→c

(f(x)

g(x)

)=

(limx→c f(x)

)(limx→c g(x)

) (when M 6= 0).

5. limx→c

(f(x)

)q= Lq (where q > 0 is rational).

From these rules we have that the limits of polyno-mials are found be evaluating the polynomial at thelimit point. Similarly for ratio of polynomials if thedenominator is not 0.

Squeeze TheoremOne way to find a limit of a function that we do not

understand is to put it between two functions that wedo understand that come together. In particular if wehave

`(x) 6 f(x) 6 u(x)

for x near c and

limx→c

`(x) = limx→c

u(x) = K then limx→c

f(x) = K.

Related to this is the observation that if f(x) 6 g(x)for x near c and the limits for both f and g exist nearc, then lim

x→cf(x) 6 lim

x→cg(x).

Algebraic manipulation of limitsWhen our limit is going to 0/0 (or possibly ∞/∞,∞−∞, 0·∞, etc.) then we have an ambiguous num-

ber since 0/0 is undefined. Most of the time whenwe encounter this we will try to manipulate what weare taking the limit of, the goal being to “cancel the0s” (i.e., rewrite it in such a way that we can cancela common term from top and bottom so that whatremains does not go to 0/0). There are three maintechniques we can use.

1. Rewriting. This is usually done when we have apolynomial and we can either expand the poly-nomials out or factor (or sometimes both).

2. Multiplying by the conjugate. The conjugate ofan expression a − b is a + b. So if we multiplyboth top and bottom by the conjugate of a − band then multiply out we get a2 − b2 (this canbe helpful for instance in getting rid of squareroots). The reason we have to multiply both topand bottom is so that we do not change the limit,i.e., multiplying by 1 does not change the valueof the limit.

3. Using identities. This is most commonly donewith limits involving trigonometry in which casethere are often many identities which we can useto rewrite (and hopefully cancel!) the terms.

Rigorous approach to limitsIntuitively, lim

x→cf(x) = L is saying that as x gets

close to c then f(x) gets close to L. We can makethis rigorous (and thus ensure that what we are do-ing will actually produce meaningful answers). Thekey is to understand “close”. In particular, by closewe are talking distance so we want to say that thedistances are small. The distance between numbersis found by absolute values, so we have “x gets closeto c” becomes |x − c| < δ (where δ is some numberto measure how close), and “f(x) gets close to L” be-comes |f(x) − L| < ε (where ε is again some numberto measure how close).

The key observation to make is that we want toensure that no matter how close we want f(x) to beto L we can guarantee this by ensuring that x is closeto c. Hence we have the following formal definition.

We have limx→c

f(x) = L if and only if for all

ε > 0 there is a δ > 0 so that if |x − c| < δthen |f(x) − L| < ε.

This leads to the classic game of you give me an εand I will find a δ. From this we can rigrously justifyall of the intuitive rules we mentioned earlier. Noteone useful method in doing so is the “add 0” method.

Quiz 2 problem bank

1. Find the average rate of change of the functionf(x) = x3 − 4x2 + 5x− 3 over the interval [0, 2].

2. Given that the average rate of change fory = f(x) over the interval [0, 3] is −1, theaverage rate of change over the interval [2, 3] is5, and the average rate of change over theinterval [2, 6] is 3, determine the average rate ofchange over the interval [0, 6].

3. Find the average rate of change of y = x2 fromx = a to x = b. If possible, use algebra tosimplify the expression.

4. Determine the unique c < 0 so that forf(x) = x3 − 2x the average rate of changebetween c and 1 equals the average rate ofchange between 1 and 2.

5. Find limx→2

√x+ 2− x

x− 2.

6. Find limx→1

x2 + x− 2

(x+ 2)2 − 9.

7. Find limx→1

3x4 − 5x2 − 7x3 + 11

5x5 − 3x3 + 1.

8. Find limx→π

4

cos(2x)cos x− sin x

.

9. Find limt→0

sin t sin 1t

.

10. Find limy→0

y2

2+ sin(y137).

One-sided limitsIn the definition of limits we look at what hap-

pens as x → c, but there are two ways that x canapproach c. Namely we can approach it from below(i.e., x < c) or we can approach it from above (i.e.,x > c). Sometimes it is convenient to limit ourselvesto one direction when evaluating the limit, and othertimes it might be possible to only approach from onedirection, which leads to one-sided limits.

limx→c−g(x) ↔ limit as we approach c from below

limx→c+g(x) ↔ limit as we approach c from above

These are also known respectively as the left limitand the right limit. (Beyonce implicitly mentionsone-sided limits in her song “Irreplaceable” whenshe sings “to the left, to the left. . . ”.)

As an example we have

limx→0−

|x|

x= limx→0−

−x

x= limx→0−(−1) = −1; and

limx→0+

|x|

x= limx→0+

x

x= limx→0+ 1 = 1.

Note that by limiting ourselves to one side we cansimplify expressions, e.g., dropping the absolutevalue signs. When the left and right limits do notagree then the limit does not exist, conversely if bothone-sided limits exist and agree then the limit ex-ists. In general, when dealing with piece-wise func-tions (such as |x|) it is convenient to use one-sidedlimits to determine what happens at the glue point.Note some limits are easier handled when treated asa combination of one-sided limits.

The limit of (sin θ)/θ as θ→ 0

A limit which plays an important role is

limθ→0

sin θθ

= 1,

where θ is measured in radians. This is done by not-ing that for 0 < θ < 1

2π that

cos θ <sin θθ

< 1

then using the squeezing theorem. This handlesthe limit from above, then by symmetry (function iseven) the limit exists and equals 1.

Note that the limit is nice and clean for radians;if θ is in degrees then the limit goes to π/180. Thisis the reason Calculus people work in radians (1 ismuch easier to work with)!

Continuous functionsClosely related to the idea of limits is the idea of

a continuous function. A function is continuous if ithas “no breaks”. Another way to say it is the function

is continuous at x = c if what we expect to happen atx = c is what actually does happen, i.e.,

limx→c f(x) = f(c).

In particular, three things need to happen to be con-tinuous: (1) f(c) must be defined; (2) the limit mustexist; (3) the preceding two have to agree. There areseveral types of discontinuities.

• Removable discontinuity The limit exists but ei-ther the function is not defined or the value ofthe function does not match the limit. The nameremovable comes from the idea that we can re-define the function at the point and we wouldno longer have a discontinuity.

• Jump discontinuity The left and right hand limitsexist but are not equal.

• Infinite discontinuity The left, right, or both lim-its are ±∞.

• Miscellaneous Another possibility, not named, isthat the limit does not exist (for example sin(1/x)at x = 0).

Examples of continuous functions include polyno-mials, xa (in its domain), sin x, cos x, tan x (awayfrom the vertical asymptotes), ln x and ex. Theseform the building blocks of continuous functions andthen we might ask for how can we combine continu-ous functions together. In particular we have that iff(x) and g(x) are continuous then so are f(x) + g(x),kf(x), f(x)g(x), f(x)/g(x) (when g(x) 6= 0) and f(g(x))(the composite function also denoted as (f ◦ g)(x)).

All of the functions that we will encounter are builtup using the basic building blocks and rules for com-bining. The only other thing that might happen isthat we might have a function which is defined piece-wise. When this is the case the interesting questionusually occurs at when two pieces meet.

The nice thing about a limit involving a continuousfunction is that we can plug in the point we are tak-ing the limit to into the function. If we get a numberout then we are done. If we get 0/0 or other incon-clusive forms (i.e., ∞ −∞, etc.), then that means weneed to work on the limit some more using our vari-ous techniques.

Intermediate value theoremThe intermediate value theorem states that for a

function f(x) continuous on the interval [a,b] that asthe input ranges from x = a to x = b that the outputwill include every possible value between f(a) andf(b) (possibly more). In particular there is no possi-ble jump in the output of a continuous function.

This can be used in finding roots, i.e., values wheref(x) = 0. If we know f(a) > 0 and f(b) < 0 and f iscontinuous in an interval containing a and b then

there must be at least one (possibly several) point be-tween a and b which is a root. Successively cuttingintervals that contain a 0 in half we can quickly findapproximations to roots. We will see other methodsto find roots of functions later in the course.

Infinite limitsOccasionally our limiting behavior involves infin-

ity (∞). Infinity is not itself a number but rather anindication of what happens when we grow in an un-bounded way. Limits involve infinity in two ways,namely we can look at what happens when the inputgets arbitrarily large (i.e., x → ±∞), or what hap-pens when the function grows arbitrarily large (i.e.,f(x)→ ±∞).

A function has a horizontal asymptote as x → ∞if limx→∞ f(x) = L where L is finite (similarly for

x → −∞). A functions has a vertical asymptote iflimx→a f(x) = ∞ (or −∞); in the latter case we often

deal with one sided limits as the two different sidescan have quite different behaviors.

To evaluate limits going to infinity it helps to di-vide out by the part which is growing fastest an seewhat happens.

Manipulating namesAn important technique in limits (and in mathe-

matics in general) is the ability to modify our limitsby changing how we represent the limit. In essencewe are substituting in another name for our limit,with the goal of being to result in an easier integral.

As an example, consider limx→∞x sin

(1x

). We can

start by letting u = 1x

, or equivalently by lettingx = 1

u. We have to now replace every occurrence

of x and these are in two places, the function weare taking the limit of, and the limiting point. Thenew function becomes sin(u)

u(nice!); the new limit-

ing point becomes 0 approached from above (i.e., asx→∞ then u = 1

x→ 0+). We can conclude

limx→∞ x sin

(1

x

)= limu→0+

sin(u)u

= 1.

Quiz 3 problem bank

1. Find limθ→0

sin(4θ) + sin(5θ)sin(θ) + sin(2θ)

.

2. Find limx→0√x+ 1− 1

sin x.

3. Find limt→0

(1

3t−

1

t(t+ 3)

).

4. Let f(x) be piece-wise defined by

f(x) =

{5x+ 3 if x 6 1,

3x− 5 if 1 < x.

Determine limx→0 f(1+ x2).

5. Let g(x) be piece-wise defined by

g(x) =

1− x2 if x < 0,

2+ 12x if 0 6 x 6 2

316x3 if 2 < x.

Determine limx→0g(x)g(x+ 2).

6. Determine a and b so that the followingpiece-wise function is continuous everywhere.

h(x) =

6+ x if x < −1

ax2 + bx if − 1 6 x 6 1

5− 6x if 1 < x

7. Find limx→9

3−√x

27−√x3

.

8. Find limx→∞

(√x2 + 5x−

√x2 − 3x

).

9. Find limx→−∞ 6x+ 4√

4x2 − x− 5.

10. Given that limx→1g(x) exists and

limx→1

(1

g(x) − 5−

1

g(x) + 3

)=1

6,

determine the possible value(s) of limx→1g(x).

Tangent linesNow that we have limits we see that the slope of

the tangent line at x = a can be found by taking thelimit of slopes of secant lines between a and a pointthat is moving closer to a. We call the slope of thetangent line at x = a and denote it by f ′(a). So wehave

f ′(a) = limh→0

f(a+ h) − f(a)

h= lim

b→a

f(b) − f(a)

b− a.

(The second definition is an alternative definition,but is based on the same idea where we take the limitof slopes of secant lines.) We note that if f(x) and xhave units then the units of f ′(a) are (the units off)/(the units of x); as an example if f measures costin dollars and x measures number of items then theunits of the derivative would be (dollars)/(item) (or“dollars per item”).

Once we have the slope of the tangent line it is easyto find the tangent line (since we also have the point(a, f(a)

). Namely, since in general a line has the form

y− y0 = m(x− x0)

then we can substitute (x0,y0) =(a, f(a)

)and m =

f ′(a) to get

y− f(a) = f ′(a)(x− a) or y = f(a) + f ′(a)(x− a).

Of course we can also recover information if wehave the tangent line. For instance if we know thetangent line and at what value of x = a it is tangentat we can recover f(a) and f ′(a) (by looking at the ycoordinate of the tangent line at x = a and the sloperespectively). But it is also important to note that thisis all the information that we know about f(x), i.e., wecannot say anything about f for x away from a.

Problem solving adviceSolving problems (and in Calculus we have many

problems) goes through four stages. It is useful toconsciously think about these stages, and eventuallyto have them become automatic for us.

1. Understand the problem. Make sure to identifywhat we are given, what we are trying to find,and what connects them. Understand all of thewords and symbols. Are there similar problemswe have tried in the past?

2. Make a plan. Outline the steps that need to becarried out. Understand what pieces of the prob-lem there are and what order things need to besolved in. Don’t be intimidated by problems thatlook big and complex, problems break downinto smaller and simple pieces.

3. Carry out the plan. The important thing is to fol-low through the steps carefully, small mistakes

snowball into big mistakes. As you carry out theplans be responsive to modifying the plan, per-haps a better method will reveal itself.

4. Look back. Is your answer reasonable? Can youcheck it by some other method? What were theindicators that will help you identify the propermethod to find a solution in the future?

The best problem solving advice: Try something. Ifit doesn’t work, try something else. But never give up!

Test taking adviceWhen taking tests, don’t get too hung up on a

problem. Go until you are stuck, think for a minutethen go to another problem. When you return youwill find that the problem has gotten easier and nowyou can make more progress.

Make sure to give yourself time to review yourwork for errors (as much as possible). If tests allowpartial credit do your best to answer as much as youcan. In general, make sure to grab the low hangingfruit, do the easy problems first! If something seemsimpossible or computations are becoming ridiculous,stop and think for a minute, usually this is a sign amistake has been made or that there’s a better way.

Quiz 4 problem bank

1. Given that y = 3x+ 4 is tangent to y = f(x) atx = 2, determine f(2) and f ′(2).

2. Given that y = 5x− 3 is tangent to y = g(x),determine a tangent line to y = 2g(x).

3. Given that y = −2x+ 4 is tangent to y = h(x),determine a tangent line to y = h(x+ 2).

4. Given that y = 4x− 7 is tangent to y = f(x) andthat f(x) is invertible, determine a tangent lineto y = f−1(x).

5. Determine c so that for f(x) = x2 the averagerate of change between a and b equals theinstantaneous rate of change at c.

6. Find the instantaneous rate of change fory = (sin x− cos x)2 + sin(2x) at x = 2

3π.

7. Find the value a so that the tangent line to

y =1√x

at x = a is parallel to the line

y = −4x− 137.

8. Find all lines of the form y = k which aretangent to the curve y = x2 − 4x+ 3.

9. Find all lines of the form y = kx which aretangent to the curve y = x2 − 3x+ 4.

10. Find a and b so that y = ax2 + bx+ 5 is tangentto the line y = 2x− 3 at (2, 1).

Notation for derivativesWe can think of the derivative f ′(x) as a function,

where evaluating the function at any point gives usthe slope of the tangent line to the curve y = f(x). Ingeneral we have

d

dx

(f(x)

)= f ′(x) = lim

h→0

f(x+ h) − f(x)

h.

(The notation “d/dx” comes from Leibniz and shouldbe read as “the derivative with respect to x”.)

We can denote the derivative at a particular pointa by f ′(a), or using the Leibniz notation we have

d

dx

(f(x)

)∣∣∣∣x=a

(the “|” indicates that we are evaluating, we will seethis notation later when we get to integration).

If the derivative exists, we note that for small h thatf(x + h) − f(x) ≈ f ′(x)h. One consequence of this isthat f(x+h)−f(x) must be small and so we have thatif the derivative exists at a point then the functionmust be continuous at that point. Conversely, if thefunction is not continuous then we automatically canconclude that there is no derivative.

Rules for derivativesActually using the limit definition to calculate the

derivative is tedious and can get quickly compli-cated. So we will build up a collection of rules to al-low us to calculate the derivatives of functions. Hereare a few basic ones to get us started:

1.d

dx

(1)= 0.

2.d

dx

(x)= 1.

3.d

dx

(xa

)= axa−1.

4.d

dx

(ex) = ex.

5.d

dx

(f(x) + g(x)

)= f ′(x) + g ′(x).

6.d

dx

(kf(x)

)= kf ′(x).

7.d

dx

(f(x)g(x)

)= f ′(x)g(x) + f(x)g ′(x).

8.d

dx

(f(x)

g(x)

)=f ′(x)g(x) − f(x)g ′(x)(

g(x))2 .

The first three rules tell us how to take the deriva-tive of xa, the basic idea is to bring the power of xdown in front and then subtract 1 from the power(note if you ever encounter a

√x or similar term it

is best to rewrite is as x1/2 before applying the rule).

The fourth rule is for the differentiation of the expo-nential function. The remaining rules give us waysto take the derivative of combinations of functions.For instance the fourth rule says if we have two func-tions being added (or subtracted) then to take thederivative we take the derivative of each piece andadd (or subtract) the result. The fifth rule says if wehave a constant times a function the derivative is thesame constant times the derivative of the function.The sixth rule tells us how to take the derivative ofa product (hence it is known as the product rule andis one of the most important results we will learn inthis class), similarly the seventh rule tells us how totake the derivative of a quotient (hence it is knownas the quotient rule it is not one of the most importantresults we will learn in this class, but still good toknow.

There is one other way to combine functions andthat is by composition. This leads to a rule for deriva-tives known as the chain rule, but we have not cov-ered that and it will appear on a future quiz.

Once we have the derivatives of the basic functionsand the rules for how to take derivatives of variousways of combining functions we will be able to takederivatives of all the functions that we encounter.That is why it is important for us to learn these rules!

Using the derivativeThe reason that we are going through all of this is

that we can use the derivative of a function to giveus information about the function. For instance, byknowing the slopes of the tangent lines we can knowwhether the function is increasing or decreasing (i.e.,where the function is going “up” the slopes of thetangent lines, or the derivative, is positive and wherethe function is going “down” the slopes of the tan-gent lines, or the derivative, is negative). This helpsus to graph a curve and let us identify importantfeatures on the curve (we will discuss these ideas inmore detail later).

Derivatives also show up in economic analysis. Forexample in economics we are often concerned withthe cost of manufacturing goods. An important mea-surement that comes up in analysis is the marginalcost which corresponds to the cost of producing thenext item. So for example we have

C(q+ 1) − C(q)︸ ︷︷ ︸= marginal cost

=C(q+ 1) − C(q)

(q+ 1) − q≈ C ′(q),

i.e., the marginal cost can be interpreted as a slope ofa line between

(q,C(q)

)and

(q+1,C(q+1)

)which is

approximately the slope of the tangent line. In prac-tice it is more convenient to use the derivative as thedefinition of the marginal cost in doing the analy-sis and answering questions about that is happening.Note that a similar thing happens with the marginalrevenue.

Derivatives also arise frequently in physics. Be-fore we discuss an example let us introduce higherorder derivatives. Since f ′(x) is a function of x wecan take its derivative, we get a function denotedf ′′(x) which is the second derivative of x. Since f ′(x)tells us something about how the function is chang-ing (via slopes of tangent lines), then f ′′(x) also tellsus something about how the function looks by see-ing how the slopes of the tangent lines are chang-ing. (This last idea is known as concavity and wewill come back to this later.)

Of course, we can take the derivative of the sec-ond derivative and get what is known as the thirdderivative; and yet another derivative would give usthe fourth derivative and so on. In general, the nthderivative of f(x) is denoted by f(n)(x) or in Leibniznotation dn

(f(x)

)/dxn.

As an example of this, suppose that we have afunction s(t) which measures distance in some units.Then the first derivative, s ′(t). is how position ischanging and is called the velocity (velocity is differ-ent from speed in that velocity has a sign (alterna-tively, an orientation), i.e., speed is |s ′(t)|), while thesecond derivative, s ′′(t) is how the velocity is chang-ing and is called the acceleration.

Quiz 5 problem bank

1. Findd

dt

(t3 − 2t2 + 2t− 4

t2 + 2

).

2. Determine the value x for whichf(x) = x3 − 9x2 + 5x− e7 has the lowestinstantaneous rate of change; also give theinstantaneous rate of change at that point.

3. Two curves passing through the same pointintersect perpendicularly if their tangent lines areperpendicular. Choose a and b so that y = 1

eex

and y = ax2 + bx intersect perpendicularly atthe point (1, 1).

4. Expressd2

dx2

(f(x)g(x)

)in terms of f(x), f ′(x),

f ′′(x), g(x), g ′(x) and g ′′(x).

5. Expressd

dx

(f(x)g(x)h(x)

)in terms of f(x), g(x),

h(x), f ′(x), g ′(x) and h ′(x).

6. Given that f(1) = 2, f ′(1) = −1, g(1) = 3 andg ′(1) = 2 find h(1) and h ′(1) whereh(x) = x2f(x) − 3

√xg(x).

7. Given that y = 3x− 7 is tangent to y = f(x) atx = 2, find the tangent line to y = xf(x) at x = 2.

8. Given that y = 51x− 117 is tangent toy = x2g(x) at x = 3, find the tangent line toy = g(x) at x = 3.

9. Newton’s first law states that an object inmotion tends to stay in motion. Suppose aparticle is being moved along the curvey = 8+ 4ex − 7x for x < 0. At time x = 0 theoutside force stops acting on the particle and itcontinues in its current motion. Determine the(positive) location where the particle will hit thex-axis.

10. Given that the distance of a particle is given bys(t) = t4 − 24t2 + π17t+ e3 + 6, determine allvalues t so that the acceleration of the particle iszero.

Derivatives of trigonometric functionsUsing the limit of sin(h)/h as h→ 0 and some ba-

sic trigonometric identities we are able to find deriva-tives for the trigonometric functions.

d

dx

(sin x

)= cos x

d

dx

(cos x

)= − sin x

d

dx

(tan x

)= sec2 x

d

dx

(sec x

)= sec x tan x

d

dx

(csc x

)= − csc x cot x

d

dx

(cot x

)= − csc2 x

(The last two of these are less commonly used.)The important thing to remember is to get the signs

on the derivative of sine and cosine correct. Since wecan now be tested on these derivatives it is also help-ful to know some of the values of the trigonometricfunctions.

θ 0 16π 1

4π 1

3π 1

2π π

sin θ 0 12

12

√2 1

2

√3 1 0

cos θ 1 12

√3 1

2

√2 1

20 −1

tan θ 0 13

√3 1

√3 DNE 0

Chain ruleThe chain rule deals with the situation of how to

take the derivative of a function inside of anotherfunction. Often a hint that we will use the chain ruleis parentheses “(” and “)” or if when reading it outloud we use the word “of”. For example sin(x2) (read“sine of x squared”) is the function x2 inside the sinefunction.

d

dx

(f(g(x)

))= f ′

(g(x)

)g ′(x)

We can of course go several layers deep, i.e., a func-tion within a function within another function. Inthis way math is like onions, there are lots of layersand occasionally a few tears.

Common mistakes when using the chain rule in-clude forgetting to multiply by g ′(x) or puttingf ′(g ′(x)). So for example

d

dx

(sin(x2)

)= cos(x2)·2x

and not cos(2x) or cos(x2). Also make sure you haveclearly identified what the inside and outside func-tions are, i.e., sin(x2) is a very different function than(sin x)2.

An important application of the chain rule is thepower rule, which is the special case when f(x) = xa.

d

dx

((g(x)

)a)= a

(g(x)

)a−1g ′(x).

Implicit differentiationAn explicit function is when you have

y = stuff with x .

An implicit function is when we are given a relation-ship involving x and y but we do not (or can not)write y as a simple function of x. For example,

yx2 + 3x cosy = sin x+ y

gives a relationship between x and y, and for a givenvalue of x we can solve for what possible value(s) ycan be. So even though y is not given as a functionof x we still can think of it as depending on x (beingthe case that we don’t know exactly which functionit is).

So to take the derivative we start with the equa-tion defining the implicit relationship and take thederivative of both sides with respect to x. There area few things to observe

1. If we take the derivative of two functions whichare equal the derivatives are still equal.

2. We treat y as a function of x, in particular whenit comes to expressions involving y we use thechain rule to take the derivative and when wefinally get to the point where we need to writedown the derivative of y we put dy/dx or y ′

(both mean the derivative of y, how easy isthat!).

3. Finally we solve for dy/dx or y ′ by simply rear-ranging the result.

We can also find higher order derivatives by re-peating this procedure.

So as an example if we differentiate both sides ofthe implicit relationship given above we have

x2y ′ + 2xy+ 3 cosy− 3x(siny)y ′ = cos x+ y ′

which we then put terms with y ′ on one side andterms without y ′ on the other side and then factorout y ′ and divide by its coefficient. In this case weend up with

y ′ =cos x− 2xy− 3 cosyx2 − 3x siny− 1

.

And we can use this for example to see at (0, 0) thatthe implicitly defined curve has slope 3, i.e.,

y ′∣∣∣∣(0,0)

=1− 0− 3

0− 0− 1= 2.

(The “∣∣(a,b)” is notation that stands for evaluate at

the point (a,b).)One easy mistake is to forget the y ′ terms in

implicit differentiation, or forgetting to differentiateboth sides (particularly common when one side is aconstant!).

Quiz 6 problem bank

1. Find the tangent line to y = sin(πx2) − πx atx = 1.

2. Given f(t) = sin t cos t, find f(101)(t).

3. Determine a and b so that f(x) will have aderivative at x = 0, where f(x) is definedpiecewise by

f(x) =

{3 tan x− 2 sec x if x < 0,

ae2x + bx if x > 0.

4. Consider the functions f(x) = 2 tan x+ 2 sec xand g(x) = (tan x+ sec x)2 + 1. Is g(x) thederivative of f(x) or is f(x) the derivative ofg(x)? Justify your answer.

5. Findd

dx

(sin

(ex + cos

(x2 + x tan x

)))6. Given that y = 4x+ 1 is tangent to y = f(x) atx = 1, and that g(x) = 3f(x2) − x3, thendetermine two tangent lines to y = g(x).

7. Let f be a twice differentiable function (i.e., hasa second derivative) which satisfies thefollowing:

f(2)=−1 f(3)= 5 f(4)= 2f ′(2)= 3 f ′(3)= 2 f ′(4)=−3f ′′(2)= 2 f ′′(3)= 1 f ′′(4)= 4

Determine g ′′(1) where g(x) = x2f(x2 + 2x).

8. An implicit function has a vertical tangent line ifdx/dy = 0. Find the four points on the implicitlydefined function 2y3 + 3y2 = 4x2 + 5x+ 1where we have a vertical tangent line.

9. Find the tangent line to the implicitly definedcurve y3 + 3x2y+ x3 = 5 at (1, 1).

10. Findd2y

dx2given that y+ 1

3y3 = x+ 137.

Derivatives of inverse functionsGraphically a function is related to its inverse by

flipping across the line y = x (i.e., interchanging theroles of x and y. In particular a tangent line to f(x)at

(a, f(a)

)will when flipped across the line y = x

will become a tangent line to f−1(x) at(f(a),a

). We

can make this precise by using implicit differentia-tion. Namely if y = f−1(x) then we have x = f(y),taking the derivative of both sides have 1 = f ′(y)dy

dx,

or rearranging:

d

dx

(f−1(x)

)=

1

f ′(y)=

1

f ′(f−1(x)

)So when there is a function which has a derivative,

we can now find the derivative of the inverse func-tion. As a simple example, take the exponential func-tion f(x) = ex which has as its inverse f−1(x) = ln x.Since the derivative of ex is again ex we have

d

dx

(ln x

)=

1

elnx =1

x.

More generally:

d

dx

(ln |x|

)=1

x

Combining these results with the chain rule and basicalgebra we also have the following:

• d

dx

(ax

)= ax lna

• d

dx

(loga x

)=

1

x lna

Logarithms are very useful, they satisfy severalnice properties, i.e., ln(ab) = lna+ lnb and ln(ab) =b lna. This allows us to take a complicated expres-sion composed by multiplying multiple functionsand/or raising functions to a function and then tak-ing a log to simplify expressions. This is known aslogarithmic differentiation. As a simple example con-sider y = xx. First we take the log of both sidesgiving lny = x ln x and now we take derivativesof both sides, i.e., y′

y= ln x + x1

x= ln x + 1 or

y ′ = y(ln x+ 1) = xx(ln x+ 1).

Inverse trig functionsAmong the inverse functions that will prove use-

ful are the inverse trig functions (sometimes denotedwith “arc”). The key is to use various trig identitiesto rewrite our expressions. As an example (a per-sonal favorite!) if y = arctan x then x = tany or tak-ing the implicit derivative we have 1 = (sec2 y)y ′ ory ′ = 1/ sec2 y = 1/(tan2 y + 1) = 1/(x2 + 1). Similaranalysis gives the other inverse trig functions and sowe have the following.

1.d

dx

(arctan x) =

1

1+ x2

2.d

dx

(arcsin x) =

1√1− x2

3.d

dx

(arcsec x) =

1

x√x2 − 1

Related ratesDerivatives measure rate of change. In many prob-

lems it often occurs that the change of one variablewill be connected with the change of another vari-able; in particular if we know how one of the vari-ables is changing we should be able to say somethingabout how the other variable is changing since theirrates of change are related. Related rate problemsare usually easy to identify since they will give a rateand ask for another rate.

There is a slight twist compared to what we didpreviously in that we usually think of how somethingchanges with time and so instead of taking a deriva-tive with respect to one of our variables we will al-most always take it with respect to time t (or somesimilarly appropriate variable), so it is important tohave the chain rule and to apply it correctly.

In general there are a few simple steps to solve aproblem about related rates.

1. Find a relationship (i.e., an equation) betweenthe variables that are changing. It often helps todraw a picture if one is possible. There are only afew handful of variations that we will encounter.Most involve either the Pythagorean Theorem,similar triangles or areas and volumes of basicshapes. (You will be expected to know the areaof rectangles and circles and triangles, and thevolumes of boxes and cylinders.)

2. Take the derivative (of both sides) with respectto t to get a relationship for the rates.

3. Plug in all the values that you know to find thevalue that you are looking for.

The hardest part about these problems is almostalways finding the relationship. Another thing towatch out for is they may only give the value for onevariable, in which case you might need to solve forthe value of the other variable using the relationship.Finally, an important tool in solving this type of prob-lem is being able to strip out all the unnecessary in-formation and translating a word problem into some-thing like a calculus problem (“math-a-nese”).

A classic example is a 10 foot ladder that is slidingdown a wall and you notice that when the bottomof the ladder is 6 feet from the wall that it is slidingaway from the wall at a rate of 1 foot per second.How fast is it sliding down the wall? To answer this

we can draw a simple picture and see that the ladderthe floor and the wall make a right triangle with afixed hypotenuse of length 10. If we let x denote thedistance of the bottom of the ladder to the wall andy the height of the ladder then we have x2 + y2 =100 (note that when x = 6 it is easy to see from thisrelationship that y = 8). The fact that the ladder ismoving away from the wall tells us that dx/dt = 1and that we are after dy/dt. Taking the derivativeof both sides we have 2xdx

dt+ 2ydy

dt= 0. Plugging

in what we know we have 2·6·1 + 2·8·dydt

= 0 so thatdydt

= −34

feet per second.Note that units will always do what you think they

should do. And so we do not need to keep track ofthem in our calculations if we know what the endunits should be, they will work themselves out cor-rectly!

Quiz 7 problem bank

1. Findd

dx

(ln

(ln

(ln |x|

))).

2. Findd

dx

((1+ x2)sinx

e5x+ 3x

).

3. Given that f(x) = x3 + ex−1, and thatg(x) = f−1(x), find g ′(2).

4. Findd

dθ

(ln(sec θ+ tan θ)

), simplify the answer

as much as possible.

5. Findd

dx

(arctan x− 1

2arctan

(12

(x− 1

x

)))for

x 6= 0, simplify the answer as much as possible.

6. From a small island in the middle of a largelake you set out in a canoe and head east, and ashort time later your friend sets out in a canoeand heads north. After a few hours your friendcalls you on your walkie-talkie to see how faryou have gone. You respond “I am fine, buthaven’t kept track of distance but right now Iam going at a speed of one mile per hour.”Your friend comments back, “That’s prettygood, I have gone a few miles and right nowam only doing one-third of a mile per hour.”Glancing at your walkie-talkie you see that it isindicating that you and your friend arecurrently five miles apart and are moving apartfrom each other at a speed of one mile per hour.Find the distances that you and your friendhave traveled in the canoe.

7. While studying for the quiz some studentsdecide to take a break and bake brownies. Theystir the brownie batter in a bowl which has ahemispherical shape with a radius of 5 inches,they then pour it into the only pan they have

which is a large circular cake pan (i.e., theshape of a cylinder) which is 12 inches indiameter and 2 inches deep. One studentobserves that the depth of the batter in the panraised at a constant rate of 1/6 of an inch persecond. Find the rate at which the depth of thebatter is falling in the mixing bowl when thedepth of the batter in the mixing bowl is 3inches. (Hint: the volume of batter of depth h ina hemispherical bowl with radius r isπ(rh2 − 1

3h3).)

8. Starfleet intelligence has recently learned of anew threat. A new Borg vessel has beendiscovered that can change its shape, they arecalling it the B-1000 (short for Borg-1000). TheB-1000 has some limitations, first the only shapeit can have is a three dimensional box andsecond the volume is always fixed, i.e., the boxcannot deviate from a fixed volume. From anearlier observation you saw that the B-1000 haddimensions 20 meters by 15 meters by 10meters. Currently though you can only see twosides. You notice that the length is currently 12meters and is increasing at a rate of 1 meter perminute; the width is currently 25 meters and isdecreasing at a rate of 2 meters per minute.What is the current depth of the B-1000 andhow fast is it changing?

9. While visiting family you attend a pumpkindrop, an event where a very large pumpkin islifted by a crane to a height and then droppedresulting in a beautiful moment of pumpkinexplosion. You pull out your camera to recordthe pumpkin from the moment it is releasedand follow it down until it hits the ground. Youhappen to notice that you are standing 32 feetaway from where the pumpkin will land, andyou also notice that the pumpkin is being liftedto a height of 64 feet before being dropped.Knowing your basic physics you determine thepumpkin’s height at a given time t will be64− 16t2 where t is the number of seconds afterthe pumpkin is dropped. If θ measures theangle of elevation your camera makes with theground, what is dθ

dtat the moment the pumpkin

will hit the ground?

10. A particle moves along the curve implicitlydefined by xy4 − yx4 = x− y2. When theparticle passes through the point (1, 1) its xcoordinate is changing 1/4 units per second.How fast is the y coordinate changing?

Linear approximationSometimes we do not need the exact value of a

function (or it might be very hard to compute) but anestimate will do. In that case we can use the tangentlines at a point that we understand what is going onto get an estimate at a point nearby that we want toknow what is going on. (Note a key word to look forin a problem is “estimate”, if we see it then we knowwe are doing some form of linear approximation.)

Recalling one of the definitions of the derivativewe have that at a point a that

f ′(a) = limx→a f(x) − f(a)x− a

= limx→a ∆f∆x ,

where ∆f is a small (but measurable!) change in thefunction and similarly ∆x is a small change in theinput. In particular we have that if x is close to a, i.e.,∆x is small, then

f ′(a) ≈ ∆f

∆xor ∆f ≈ f ′(a)∆x.

One example of where this might come in is if wewant to estimate the result of a small change in xwill have on the change in f or vice-versa. It also isuseful for getting an estimate on how much error inour input will effect the error in our output.

If we replace ∆f by f(x)− f(a) and ∆x by x−a andrearrange then we have

f(x) ≈ f(a) + f ′(a)(x− a)︸ ︷︷ ︸=L(x)

.

The right hand side, the L(x) also known as the locallinearization, is the same as the tangent line at x =a and this states that the value of the function forx near a is approximately the same as the value ofthe tangent line. We can use this if we know thevalue of the function somewhere near x and want anestimate for the function at x. For instance supposewe want an estimate for

√65, we know

√64 = 8 and

65 is near 64 and so we will approximate the functionf(x) =

√x around a = 64 using the tangent line and

use that to estimate√65. So for example we have

f ′(x) = 1/(2√x) and so for x near 64 we have

√x ≈ f(64) + f ′(64)(x− 64) = 8+ 1

16(x− 64).

So we can conclude that√65 ≈ 8 + 1

16= 8.0625 (the

actual value is 8.06225 . . .).Sometimes these relationships are expressed using

differential notation, i.e.,

dy = f ′(x)dx

where dy and dx are the differentials in y and x.

Global extrema

Extrema are maximums (the largest possible value)and minimums (the smallest possible value) of thefunction. There are two types of extremums. One isglobal extremum which says that you are either thelargest or smallest possible value for all x we are con-sidering. The other is local extremum which says youare either the largest or smallest possible value for allx nearby (i.e., if we zoom in close enough and ignoreeverything else it is the largest or smallest possiblevalue). Note that a global extremum is also a lo-cal extremum but a local extremum might not be aglobal extremum.

Extreme Value Theorem: If a function is continuouson a closed interval (an interval which includes theendpoints) then the function has a global maximumand a global minimum on the interval.

Knowing that there is an extreme value is useful,but we still have to find where it is. One importantobservation is that we can use the derivative to helpus locate where the extremums might occur. This isbecause the derivative tells us locally what our func-tion is doing and we can use that information to ruleout points. For example we have the following.

f ′ > 0 ←→ f is increasingf ′ < 0 ←→ f is decreasing

But if we are at a maximum or a minimum then weare neither increasing or decreasing (if we were wewould move slightly and be even more maximer orminimer but those aren’t even words so of course thisis not possible). So we can conclude that if we are atan extremum that we must be at a point where oneof three things happens: (i) f ′(c) = 0; (ii) f ′(c) isundefined; (iii) we are at a boundary (and so unableto move in one direction). We call these points criticalpoints.

While extremum will happen at critical points, justbecause we are at a critical point does not mean weare at a minimum or a maximum. For example thefunctions y = x3 and y = 3

√x both have critical points

at 0, but 0 is not an extremum for either function.So our method to find Global extrema is as follows:

1. Make a list of x values where the global extremacan occur.

(a) Endpoints of the interval.

(b) The derivative is 0.

(c) The derivative is undefined.

2. Plug each of the above values into the function.Largest answer is the global max; smallest an-swer is the global min.

3. Profit!

Note that while we only have one global maximumit can occur at multiple points. For instance y = sin xon the interval 0 6 x 6 4π has a global maximum of1 and it occurs at two points, 1

2π and 5

2π.

Rolle’s Theorem and Mean Value TheoremRolle’s Theorem states that if a function is contin-

uous on an interval a 6 x 6 b, differentiable fora < x < b and f(a) = f(b) then there is some point cin between a and b (i.e., a < c < b) so that f ′(c) = 0.

The Mean Value Theorem takes Rolle’s Theorem andturns it slightly. Namely, if a function is continuouson an interval a 6 x 6 b and differentiable for a <x < b then there is some point c between a and b(i.e., a < c < b) so that

f ′(c) =f(b) − f(a)

b− a.

(The right hand side is the slope of the secant linethrough

(a, f(a)

)and

(b, f(b)

)and also the average

rate of change between a and b while the left handside is the slope of the tangent line at c or the in-stantaneous rate of change. So the theorem says thatif our function is “smooth” then at some point be-tween a and b our instantaneous rate of change ex-actly matched the average rate of change.)

An important corollary to the Mean Value Theo-rem is that if a function has a derivative of 0 on aninterval then the function is constant on that interval.(Previously we had seen that the derivative of a con-stant was 0, this says the opposite is also true.) Inparticular if two functions have the same derivativethen they must differ by a constant.

Quiz 8 problem bank

1. For y = 5x3 − 4e3x−6, if x = 2± 0.05, uselinearization to estimate the correspondingrange for y.

2. You have recently been hired as the chiefarchitect for one of the pyramids beingconstructed by Pharaoh Sneferu in Egypt. Aftersome consultation the pharaoh has agreed to apyramid design that is 500 cubits wide and 300cubits high (a cubit is the system ofmeasurement used in ancient Egypt). Thevolume of a pyramid is 1

3b2h where b is the

length of one side of the base and h is theheight; so that the pyramid will require25, 000, 000 cubic cubits of stone. After gettingin touch with your stone contractor youdiscover that there are only 23, 000, 000 cubiccubits of stone available. The pharaoh gives thego ahead to build a (slightly) smaller pyramid,but with the same proportions as before.Estimate how many cubits smaller the base ofthe pyramid will end up being.

3. Use linearization to give an estimate for 3√1018.

4. Use the following information to get anestimate for g(f(2.1)).

x 0 1 2 3

f(x) −1 3 0 2

f ′(x) 1 −2 3 0

g(x) 3 0 1 2

g ′(x) 1 3 2 2

5. You and a classmate are preparing to give apresentation in your astronomy course. Youhave decided that the best way to show how astar gets sucked into a black hole is through amodern interpretive dance where you will beplaying the part of a large blue class O star andyour partner will be the black hole. You willrepresent these two astronomical features usingpaper mache, and you are responsible formaking your star. Initially you were planning toblow up a spherical balloon to a diameter of 16inches before covering it in paper mache, but inyour excitement you ended up blowing theballoon to a diameter of 17 inches. Using linearapproximation get an estimate of how muchmore surface area the 17 inch balloon has ascompared to the 16 inch balloon (i.e., anestimate of how much more paper you willneed to make your model). (Hint: the surfacearea of a sphere of radius r is 4πr2.)

6. List all critical points ofg(x) = x2/3

(5x2 − 8x− 40

)+ 137eπ.

7. List all critical points of f(t) = t√4et − (t+ 2)2.

8. Find the global maximum and global minimumfor h(x) = x2 − 2 arctan(x2) for 0 6 x 6 4

√3.

(Hint: h( 4√3) =

√3− 2

3π ≈ −0.3623.)

9. Consider the following function which iscontinuous and differentiable for all x (you donot need to prove this!):

f(x) =

{x3 + 2x2 − 4x+ 2 if x 6 1;

3x− 2 if x > 1.

For the interval −4 6 x 6 2 find all values of cthat satisfies the Mean Value Theorem. (Hint:f(−4) = −14.)

10. Find the unique value c that satisfies the MeanValue Theorem for the functionf(x) = arctan(sin x) on the interval 0 6 x 6 π/2.

Local extremum and the first derivative testWhen we are looking for local extremum then we

look at the critical points (and endpoints if we havethem, but we won’t always). So we know where tolook so now the important thing is how to identifywhether a critical point is a local maximum, a localminimum or neither. One way to help distinguishbetween the possibilities is to use information abouthow the function is behaving near the critical pointby using the first derivative.

f ′ > 0 ←→ function is increasingf ′ < 0 ←→ function is decreasing

We are now ready to discuss “the first derivativetest”. We start by finding all the critical points, unde-fined points, and boundary points. We represent allof these points on the real number line. Between anypair of points the first derivative will always have thesame sign (either positive or negative); to determinethe sign just pick any point in the interval and eval-uate. There are four possibilities for how the signof the derivative behaves around a critical point andthese are shown below.

af ′ > 0 f ′ < 0

maximum

af ′ < 0 f ′ > 0

minimum

af ′ > 0 f ′ > 0

neither

af ′ < 0 f ′ < 0

neither

In addition to helping to classify critical points thisalso helps us to answer on what intervals the func-tion is increasing and on what intervals the functionis decreasing.

Concavity and the second derivative testConcavity tells us how the function is “bending”,

so that concave up is that the function is trying tobend up while concave down is that the function istrying to bend down.

Put more concretely the graph is concave up whenthe derivatives are increasing (i.e., f ′ is increasing)which corresponds to when f ′′ > 0 (i.e., a functionis increasing when the derivative is > 0). Similarlythe graph is concave down then the derivatives are de-creasing (i.e., f ′ is decreasing) which corresponds towhen f ′′ < 0 (i.e., a function is decreasing when thederivative is < 0).

A point where concavity changes is called an in-flection point. To find inflection points we take thesecond derivative; see where it is 0 or undefined andthen mark off intervals and test each interval for theconcavity. Any point where the concavity changes is

an inflection point. (This process is similar to whatwe did to use the first derivative test to find and testlocal extremum, but this is not surprising since an in-flection point also corresponds to the maximums andminimums of the first derivative.)

Since concavity tells us how the curve is “shaped”we can also use it to tell us whether a critical point isa maximum or a minimum. However there are somelimitations. First, we can only use the second deriva-tive test on a point where the first and second deriva-tives exist, and second the test might be inconclusive(unlike the first derivative test which works for bothkinds of critical points and is conclusive). We havethe following rule: If f ′(c) = 0 and f ′′(c) exists then

f ′′(c) > 0 −→ (c, f(c)

)is a local minimum;

f ′′(c) < 0 −→ (c, f(c)

)is a local maximum;

f ′′(c) = 0 −→ the test is inconclusive.

A convenient way to remember this is with the fol-lowing adorable picture.

Sketching a curveThe idea about sketching a curve is to find the in-

teresting points (i.e., critical points, inflection points,intercepts and asymptotes). Mark the points andthen connect them with appropriately shaped curvesbased on the signs of the first and second derivatives(a souped-up version of connect the dots).

OptimizationIn optimization problems we are trying to mini-

mize or maximize some value. (It is easy to spot opti-mization problems because they will be the ones thatask you to find the “largest” or “smallest” or reallyany kind of “-est” word.)

How to solve optimization problems:

1. Find appropriate labels. In particular there areessentially two things that need labels: (a) thevalue that we are trying to optimize; (b) thevalue(s) that we can vary in our optimizationproblem. It is often useful to draw a picture,if possible.

2. Find a function for what we are trying to op-timize in terms of what we can vary (this is al-ways the hardest part!). We need to get the equa-tion down to a single variable, this is done usingconstraints (i.e., relationships that the variablesmust satisfy).

3. Use techniques for finding local max/min tofind optimal values.

These problems are no different than other prob-lems where we are looking for the maximum andminimum. The only difficulty is that we are usuallynot given a function and we need to set it up (andsometimes it is buried deep inside of a word prob-lem).

For instance we might need to find the maximumarea that we can enclose with 400 feet of fencing ifthe area is a rectangle and one side does not needfencing, since it is on a river. Then what we are tryingto optimize is the area so we label it A. The area isthe length times the width of the rectangle let us labelthese as x and y (where x will be the top and bottomand y will be a single side, the other side being theriver). Now we know that A = xy, but we need toget down to a single variable. To do this we lookback and notice that we haven’t used the informationabout the 400 feet of fencing, which is a constraint.This tells us that 2x+y = 400 so we can rearrange thisto y = 400− 2x so that A = x(400− 2x) = 400x− 2x2.We now look for critical points and to do this weuse the first derivative A ′ = 400 − 4x, setting it to 0and solving we get x = 100. We can then concludethat y = 200 so that the maximum area is 100·200 =40, 000 square feet.

When doing optimization problems it is good tostop and see if your answer is reasonable. For in-stance if our answer to the previous problem wasx = −200 then we would look for a mistake. Alsomake sure that you answer the question that is asked.In this last example we are asked to find the maxi-mum area, so our answer should be an area. If wehad been asked for the dimensions then our answershould be 100×200, i.e., the dimensions, and so on.

Quiz 9 problem bank

1. On what intervals is h(x) = x2 − 2 arctan(x2)increasing and on what intervals is itdecreasing?

2. Find the two critical points of y = x2/3e−2x/3

and use the first derivative test to determine ifthey are local min’s or max’s.

3. Find the two inflection points fory = θ2 + sin(2θ) for 0 6 θ 6 π, also identify theintervals where the function is concave up andwhere the function is concave down.

4. Verify the function f(x) = cos(x3 − 2x) has acritical point at x = 0. Use the second derivativetest to determine if it is a maximum or aminimum.

5. Find and classify the location of the criticalpoints of f(x) = 2x5 − 5x4 − 10x3 + 13.

6. Find all inflection points for y = xe−6x2

, anddetermine the intervals where the function isconcave up and where the function is concavedown.

7. For y =1

1+ x2, find the value a > 0, where the

y-intercept of the tangent line at x = a ismaximal.

8. What is the area of the largest rectangle thatyou can make where the bottom edge is on thex-axis and the top two corners lie on theparabola y = 12− x2?

9. For a > 0 find the point on the curve y =√x

closest to the point (a, 0).

10. Past analysis of previous parties has led to thedevelopment of a chip index where the higherthe chip index the better the party. In particular,the chip index is N2P where N is the number ofnacho chip bags that you have and P is thenumber of potato chip bags that you have.Given that you have $18 for your chip fund anda bag of nacho chips cost $3 and a bag of potatochips cost $1, how many bags of each chipshould you buy to maximize the chip index andthus have a most awesome party.

Newton’s methodGiven a function f(x) we might be interested in

finding the roots of the function. These are values cwhere f(c) = 0. For some functions this can easilybe done (e.g., the quadratic formula), but in generalthis is a hard problem. However, if we know the ap-proximate value of the root then we can work to getan ever better approximation (i.e., we can sometimesget good estimates for our roots).

One such approach is Newton’s method whichmakes the following observation. Suppose that c isa guess for our root, then it is reasonable to use thetangent line at c as a stand-in for the function andfind when the tangent line has a zero to find a bet-ter approximation. Note that the tangent line at c isy = f(c) + f ′(c)(x − c) and when we solve this forwhen the line is zero we get our new guess:

x = c−f(c)

f ′(c)

In general we think about iterating this process sothat we have a series of guesses, x0, x1, . . . where eachone is (hopefully) better and

xn+1 = xn −f(xn)

f ′(xn).

If we are looking for a root of f(x) = x2 − 2 (hint:this is

√2), then the recursion becomes

xn+1 = xn −f(xn)

f ′(xn)= xn −

x2n − 2

2xn=1

2xn +

1

xn.

Letting x0 = 1 this gives:

x0 = 1 ≈ 1.0000

x1 =32≈ 1.5000

x2 =1712≈ 1.4166

x3 =577408≈ 1.4142

Note that this works best when we start near a root(though it might not get near the root you are lookingfor). Moreover it is possible that the process will getinto a loop or even blow up!

AntiderivativesAn antiderivative of f(x) is a function F(x) so

that F ′(x) = f(x). As an example for the functionf(x) = 2x an antiderivative is F(x) = x2, but anotherantiderivative is F(x) = x2 + 1 and in general anyfunction of the form F(x) = x2 + C for C a constantis an antiderivative of f(x), but these are the only an-tiderivatives of f(x) = 2x. In general, if F(x) is anantiderivative of f(x) then all antiderivatives are ofthe form F(x) +C. Or in other words two antideriva-tives only differ by a constant.

Notationally we will let∫f(x)dx denote the an-

tiderivative of f(x) or the “indefinite integral” of f(x).So we have that ∫

f(x)dx = F(x) + C,

where F(x) is any antiderivative of f(x).Not surprisingly, rules for derivatives become

rules for antiderivatives. So all the following rulesare found by taking rules for derivatives and rewrit-ing them in terms of rules for antiderivatives.∫ (

f(x) + g(x))dx =

∫f(x)dx+

∫g(x)dx∫

(k·f(x))dx = k∫f(x)dx∫

xa dx =1

a+ 1xa+1 + C (a 6= −1)∫

1

xdx = ln x+ C∫

ex dx = ex + C∫sin xdx = − cos x+ C∫cos xdx = sin x+ C∫

sec2 xdx = tan x+ C∫sec x tan xdx = sec x+ C∫

1

1+ x2dx = arctan x+ C

Note that the sine and cosine now have “swappedsigns” compared to the corresponding rules for theirderivatives.

For any given function there are many antideriva-tives (since again we can add an arbitrary constant);but we might be interested in finding one specific an-tiderivative. This is possible, for example, if we knowabout the value of the function at a point since thislets us determine which C is the correct one. For ex-ample suppose we know that y ′ = 4x7 − 2 sin x andthat y(0) = 4 and we want to find y. First we notethat y will be an antiderivative of y ′, i.e.,

y =

∫(4x7 − 2 sin x)dx = 4

∫x7 dx− 2

∫sin xdx

=4

8x8 + 2 cos x+ C =

1

2x8 + 2 cos x+ C.

But we also know that y(0) = 4 and so we can usethis to find C, for example we have

4 = y(0) = 0+ 2+ C or C = 2.

So the desired function is y = 12x8 + 2 cos x+ 2. This

is a simple example of what is known as an “initialvalue problem”, i.e., we know something about howthe derivative is behaving and an initial value of thefunction and we want to determine the function.

It turns out that integration can be much trickierthan differentiation. As a general rule if you havea doubt whether or not what you are doing is cor-rect then don’t! (A lot of “intuitive” things do notwork.) In fact we will spend a significant chunk ofnext semester finding ways of rewriting expressionsso that it looks like one of the ones that we have listedabove.

Quiz 10 problem bank

1. For f(x) = x3 − 3x+ 1, find an expression forthe recurrence for Newton’s method (i.e.,xn+1 =(stuff with xn)). Given x0 = 2, find x1.

2. For g(x) = x3 − 7 use Newton’s method startingwith x0 = 2 to find x1 and x2, give your answerto four decimal places.

3. Given that xn+1 =4xn

5+4

x4nwith x0 = 137,

determine the exact value that the numbers xnare approaching as n gets large.

4. Let f(θ) = cos(θ) + 2 cos(θ) sin(θ) − sin(θ), letg(θ) = 1

2(cos(θ) + sin(θ) + 1)2 and let

h(θ) = cos(θ) + sin(θ) + 12

sin(2θ). Which ofg(θ) or h(θ) an antiderivative of f(θ)? Explain.

5. Find∫e3x + 1

ex + 1dx.

6. Find∫x3 + 3

√x+ 5

xdx.

7. Find∫

1

1+ sin θdθ.

8. Find∫|t|dt.

9. Given f ′(x) =x2

1+ x2and f(1) = 4, find f(x).

10. Suppose that y ′′(t) = 2− sin(πt) and thaty(0) = −2 and y(2) = 5. What is y ′(1)?

Riemann sumsIf we want to approximate an area we can slice it

into little strips each of which can be approximatedby a rectangle; we then add up the individual rect-angles. To get a better approximation we can makethe slices “smaller”. This is the underlying idea ofRiemann sums. Given a function f(x) and an interval[a,b] we start by partitioning [a,b] up into a partitioninto pieces by first choosing points

x0 = a < x1 < x2 < · · · < xn−1 < xn = b.

These are the bases of the rectangle and the “width”of the ith rectangle is ∆xi = xi − xi−1. To find theheights we choose points ci so that xi−1 6 ci 6 xi wethen have that the “height” is f(ci). So then we havethat the Riemann sum (which is an approximationof the area under the curve y = f(x) in the interval[a,b]) is

n∑k=1

f(ck)∆xk.

Here “Σ” (sigma) is used to indicate doing a sum.(Note that sigma and sum both start with “s”, almostas if we had planned it that way!)

In general we have thatn∑

k=1

ak

is a convenient way to compress a1 + a2 + · · · + an,the “k” here is a dummy variable that starts at 1 andgoes to n (giving n terms). Because this is a sum thenthis behaves as we expect sums to behave, i.e., we canbreak sums apart, pull out constants, etc.

To indicate the slices getting smaller we let ‖P‖ =max{∆xk}. We are interested in functions where thelimit as ‖P‖ → 0 exists, we call such functions inte-grable (all continuous functions are integrable) anddenote the limit by

lim‖P‖→0

n∑k=1

f(ck)∆xk =

∫ba

f(x)dx.

The∫

sign is a stretched out “S” and indicates theidea that we are summing up little pieces. The “x” isa dummy variable and can be replaced by any othervariable, the result will be the same.∫b

a

f(x)dx =

∫ba

f(u)du =

∫ba

f(y)dy = · · · .

While our starting point is thinking of finding area,it is important to remember that the result of the in-tegration can be positive or negative, so more appro-priately it is signed area.

Note that we have used the∫

sign to indicate an-tiderivative (also called indefinite integrals), we willsoon discover that there is a nice connection with

∫ba

(also called definite integrals).

Properties of integrals

Properties of integration follow from the definitionof Riemann sums (as well as some geometric intu-ition).∫b

a

f(x)dx =

[area abovex-axis

]−

[area belowx-axis

].

We can find the values of some integrals by findingthe area is composed of combinations of triangles,rectangles and circles (right now this is the only waywe can handle integrals of the form

√r2 − x2).

If our upper and lower bound match then there isno “area” and so the integral is 0, i.e.,∫a

a

f(x)dx = 0.

Changing the order of integration changes thesign, i.e., ∫b

a

f(x)dx = −

∫ab

f(x)dx.

Integration is “linear” in the sense that we can pullconstants out as well as break it up over addition, i.e.,∫b

a

kf(x)dx = k

∫ba

f(x)dx and

∫ba

(f(x) + g(x)

)dx =

∫ba

f(x)dx+

∫ba

g(x)dx.

This for example allows us to break the problem ofintegrating several things added together into indi-vidual parts (this is especially convenient when weneed to do one technique for integrating one part anda different technique for integrating another part).

We can break the interval we are integrating intopieces (this is convenient, for example, when we havepiecewise functions), i.e.,∫b

a

f(x)dx =

∫ca

f(x)dx+

∫bc

f(x)dx.

This is true for any relationship of a,b, c. We can alsoreverse this and combine several integrals together tomake a single integral.

If f(x) 6 g(x) on [a,b] then∫ba

f(x)dx 6∫ba

g(x)dx.

In particular, if m 6 f(x) 6M on [a,b] then

m(b− a) 6∫ba

f(x)dx 6M(b− a).

Using some basic sum identities we have the fol-lowing: ∫b

a

1 dx = b− a∫ba

xdx =1

2(b2 − a2)∫b

a

x2 dx =1

3(b3 − a3)

For a function f(x) we define the average value ofthe function on the interval a 6 x 6 b by

average =1

b− a

∫ba

f(x)dx.

Quiz 11 problem bank

1. Use a Riemann sum to approximate the area

under the curve y =4x

1+ x2from x = 1 to x = 4

by using a partition with three equal-widthparts, and choosing the left hand points on eachpart of the partition.

2. Use a Riemann sum to approximate the area

under the curve of the function f(t) =10

1+ t4

for −526 t 6 5

2by using a partition with five

equal parts and choosing the center points oneach part of the partition.

3. Let h(x) be a function such that∫20

h(x)dx = 1,∫30

h(x)dx = 2,∫40

h(x)dx = 6,∫51

h(x)dx = 5, and∫52

h(x)dx = 7.

Find∫31

h(x)dx.

4. Given that∫8−7

(4f(x) + g(x)

)dx = 10 and∫8

−7

(2f(x) + 3g(x)

)dx = 0, determine∫8

−7

(5f(x) − 3g(x)

)dx.

5. Find∫5−5

sin(te−t2)dt.

6. Find∫3−1

(2|x|+ 3

)dx.

7. Find∫10

(1+

√1− x2

)2dx.

8. Find∫√2

1

√2− y2 dy.

9. Determine a so that the average value of thefunction of f(x) = 2x+ 5 for the intervala 6 x 6 a+ 2 is 11.

10. Looking over attendance records the calculusinstructor has discovered as the termprogressed that fewer students came to lecture(thus breaking their super-sized heart). Inparticular, t weeks into the semester there were150− 20t+ t2 students per lecture. Find theaverage number of students per lecture throughthe twelfth week (i.e., from t = 0 to t = 12).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connectsthe ideas of differentiation with our new idea of in-tegration. There are two parts to the FundamentalTheorem of Calculus.

(Part I) If f is continuous on [a,b] then∫ba

f(u)du = F(b) − F(a),

where F(x) is any anti-derivative of f(x).

In particular if we want to evaluate a definite in-tegral we can now do it in two steps. First, find ananti-derivative of the function. Second, evaluate thisnew function at the endpoints and take the differ-ence. This reduces the problem of integration to thatof finding an anti-derivative.

Of course finding anti-derivatives in general arenot easy! Our main technique is to work on rewritingthe function using algebraic manipulation, trigono-metric identities, or substitution (see below) so thatwe can reduce the anti-derivative to something thatwe easily recognize.

The other part of the Fundamental Theorem of Cal-culus says that integration leads to anti-derivatices.

(Part II) If f is continuous on [a,b] and

F(x) =

∫xa

f(u)du then F ′(x) = f(x).

By combining this part of the Fundamental The-orem of Calculus, the chain rule and properties ofintegrals we have the following rule:

d

dx

( ∫g(x)h(x)

f(u)du

)= f(g(x)

)g ′(x) − f

(h(x)

)h ′(x).

We see from part II of the Fundamental theoremof calculus that the function F(x) is an anti-derivativeof f(x). So we will let

∫f(x)dx (called the indefinite

integral) denote the anti-derivative of f(x). In generalwe have that ∫

f(x)dx = C+

∫xa

f(x)dx,

where C is a constant (this constant will play an im-portant role later, it is important not to forget it).

Substitution rule

Rules for derivatives become rules for integration.One of the most important rules for derivatives is thechain rule which states

d

dx

(f(g(x)

))= f ′

(g(x)

)g ′(x).

By taking the anti-derivative of each side we can con-clude∫

f ′(g(x︸︷︷︸=u

))g ′(x)dx︸ ︷︷ ︸

=du

=

∫f ′(u)du = f(u) + C

= f(g(x)

)+ C.

This is used in many problems involving integra-tion because it can help rewrite the integral in a sim-pler form. So after the substitution we might see howto proceed and then we can solve the integral and atthe end resubstitute back to get our answer in terms ofx. The indication that we should use use substitutionis to look for a function inside of a function.

With every method we have in working with solv-ing integrals the goal is always to make it simpler. Insome sense the art of integration is the art of cumu-lative simplification. It is possible that several substi-tutions might be needed. This is fine as long as youkeep track of everything.

Remember that when we are substituting that weneed to substitute for every occurrence of “x”, i.e.,we also need to make sure we substitute for the dxterm. On some integrals, in order for us to do thiswe might need to solve for x in terms of u. For ex-ample when making the substitution u =

√x then

du = 12x−1/2 dx or dx = 2

√xdu = 2udu so that the

appropriate substitution in this case is to replace thedx term by 2udu. (Note it is easy to add and divideby constants to get what we need.)

If we are dealing with a definite integral we can doone of two things. First, we do the indefinite integral,solve it to the end to get an antiderivative and thenuse the fundamental theorem of calculus to evaluateand get our answer. Alternatively, we can changethe bounds as we make our substitution (the princi-ple is again that we are replacing every occurrence ofx, and the original bounds were in terms of x, i.e.,∫ba (stuff)dx indicates we go from x = a to x = b). So

we have∫ba

f(g(x)

)g ′(x)dx =

∫g(b)g(a)

f(u)du, where u = g(x).

Applications of integration

Cumulative change: There is a connection be-tween integration and derivatives and we can use thisto answer questions about given how fast somethingis changing, what is the total. (These types of ques-tions are easy to identify since they will involve onlyone rate (where related rates involves more than one)and will ask for a total.) By the fundamental theoremof calculus we have∫ba

f ′(t)dt = f(b) − f(a) or f(b) = f(a) +

∫ba

f ′(t)dt.

The intuition is that f ′(t)dt measures the (instanta-neous) amount of change at time t and then the “

∫”

adds them all up to find the total amount of change.If v(t) = s ′(t) is velocity then∫b

a

v(t)dt = s(b) − s(a) = displacement,

and ∫ba

|v(t)|dt = total distance,

Area: We can use integration to find the area be-tween curves. If g(x) 6 f(x) on the interval [a,b] thenthe area between these curves in the interval is

Area =

∫ba

(f(x) − g(x)

)dx.

If the curves cross then find the intersection point(s)by setting f(x) = g(x) and solving for x (also donewhen no bounds are given, or when there are sev-eral curves that define the region). Once we have theintersection point(s) we split this into several piecesand work on each piece separately (this is to avoidthe problem of “signed” areas and to simplify the in-tegrals to manageable functions).

We can also integrate with respect to y, the basicidea being to take horizontal slices. In this case if wehave x = f(y) and x = g(y) with g(y) 6 f(y) on theinterval [a,b]. Then we have

Area =

∫ba

(f(y) − g(y)

)dy.

Average value: We already saw that the way tocompute the average value of f in the interval [a,b]is by

average =1

b− a

∫ba

f(x)dx.

This value is such that the rectangle with height favgand width (b− a) has the same area as

∫ba f(x)dx.

The Mean Value Theorem for Integrals states that if fis continuous on [a,b] then there is some c ∈ [a,b] sothat

f(c) =1

b− a

∫ba

f(x)dx.

(This is actually the same as the Mean Value Theoremfor derivatives, just worded differently.)

Quiz 12 problem bank

1. Given F(x) =

∫x3

2x2

√2+ 3

√t dt, find the tangent

line to F(x) at x = 2.

2. For x > 0, find F(x) =

∫x0

6|t2 − t|dt. (Hint: find

F(x) as a piecewise function according to howwe can break up the function inside theintegral.)

3. Find∫

sin(√x) sin(2

√x)√

xdx.

4. Find∫20

√t4 + 9 dt+

∫53

4√

t2 − 9 dt.

5. Find∫√

1+√xdx.

6. Let h(x) =∫2x2+x

3x2−2

1

2+ sin tdt. Determine h ′(x).

7. Reduce the following to a single integral of the

form A

∫CB

f(x)dx for some constants A,B,C.

∫50

f(x)dx−

∫33

f(x2)dx+

∫10

3f(3x)dx

−

∫40

f(12x)dx+

∫35

f(x)dx.

8. Find∫

3x√x2 + 1+ x

dx.

9. Given that∫x3

g(t)dt =3√x2 − 1+ Cx, find C

and g(x).

10. Find∫10

2x√1− x4 dx.

L’Hospital’s formula

At the beginning of the class we developed limitsto help find derivatives which involved expressionswhich went to 0

0. Now we can come full circle and

use derivatives to help us find limits that are of theform 0

0. We have the following.

(L’Hospital’s Formula) If f(x) and g(x) aredifferentiable around x = a and if as x → awe have f(x)

g(x) → 00

or → ∞∞ , then

limx→a f(x)g(x)

= limx→a f

′(x)

g ′(x).

This follows by noting that we can manipulate thelimit to a derivative. As an example we have thefollowing:

limθ→0

sin(5θ)e2θ − 1

= limθ→0

5 cos(5θ)2e2θ

=5

2.

If needed we might need to apply this severaltimes, we keep going until we can make a decisionabout the limit. We can use this to establish somebasic facts. For example we have the following, sup-pose that the function f(x) has a second derivative,then

limh→0

f(x+ h) − 2f(x) + f(x− h)

h2

= limh→0

f ′(x+ h) − f ′(x− h)

2h

= limh→0

f ′′(x+ h) + f ′′(x− h)

2= f ′′(x).

(Note in this example the variable we took the deriva-tive with respect to was h as that is what is in thelimit.)

The expressions 00

and ∞∞ are not the only timeswe cannot decide what is happening, other indeter-minate forms include ∞∞ , 0·∞, 1∞, ∞ − ∞, and soon. The typical approach is to find a way to rewritethese expressions so that it approaches either 0

0or∞∞ . One particular interesting variation is when our

limit involves an expression where both the base andexponent are changing. In this case it is helpful toobserve the following two ideas. First the logarithmfunction is continuous. Second if f(x) is continuousthen

limx→a f

(g(x)

)= f

(limx→ag(x)

).

So for example suppose we want to find

limx→∞

(1+

t

x

)x

which is approaching 1∞ we let y be our answer andthen note

ln(y) = ln(

limx→∞

(1+

t

x

)x)= limx→∞ ln

(1+

t

x

)x= limx→∞ x ln

(1+

t

x

)= limx→∞

ln(1+ t

x

)1x

Now we can apply L’Hospital and this becomes

ln(y) = limx→∞

−t/x2

1+t/x

−1/x2= limx→∞ t

1+ t/x= t.

Now we remember that we wanted y, so to finish wehave that ln(y) = t becomes y = et.

Separable differential equationsAn important part of mathematics is differential

equations which relates how a function is changing(i.e., its derivative) with the current value of the func-tion and/or the input of the function. (We have pre-viously done a very special case of this when thederivative is only in terms of the input and then re-covering the original function.)

The problem of solving differential equations isvery hard and is still being worked on. There area few special cases where we can make progress andwe will do one here (don’t worry we will get to spenda whole semester on these later). In particular weconsider separable differential equations which are ofthe form

y ′ =f(x)

g(y).

The key observation is that we can rewrite this as

g(y)y ′ = f(x).

We now think of both sides as functions of x (wait forit) and since both sides are equal there antiderivativesare also equal. The antiderivative of f(x) is a typicalantiderivative problem. For g(y)y ′ we note∫

g(y)y ′ dx =

∫g(u)du

by making the substitution u = y and du = y ′ dx,so this is also an antiderivative problem (there it is!).So taking antiderivatives of both sides they agree upto a constant and then we use additional information(i.e., some initial value) to determine the constant.Note, we can sometimes solve for y, but it is usuallybest to solve for the constant straight away (i.e., oncewe have taken antiderivatives).

This process is best seen with an example. So sup-pose we have that y ′ = − 1

10y and y(0) = 20 where y

is a function of t (this is typical of things such as ra-dioactive decay where things decrease in proportionto the amount of material). We separate by putting

all of the variables of one type on one side and all theother variables on the other

y ′

y= −

1

10integrating gives ln(y) = −

1

10t+ C.

(Note we only need a “+C” on one side, i.e., wecan combine several constants to form a single uber-constant.) Solving for y then we get

y = e−t/10+C = e−t/10eC = Det/10.

Finally to get the constant we use the initial conditionthat at time 0 we have y = 20 so this gives D = 20,

y = 20e−t/10.

Of course there are many other possibilities. Sup-pose that y ′ = x

2y+4 and y(√7) = 0. First we sepa-

rate:(2y+ 4)y ′ = x

Then we integrate:

y2 + 4y =1

2x2 + C

Then we solve for C and then y; or we solve for yand then C. In this case, let’s solve for C first. Wehave

0 =1

2

√72+ C giving C = −

7

2.

Updating we now have

y2 + 4y =1

2x2 −

7

2.

Now to solve for y we note it would be helpful tocomplete the square on the left by adding 4 giving

y2+ 4y+ 4 =1

2x2−

7

2+ 4 or (y+ 2)2 =

1

2(x2+ 1).

Finally we solve by taking the square root and sub-tracting 2 giving

y = −2+

√1

2(x2 + 1).

(Note there were actually two possibilities when wetook the square root, i.e., ±; both should be checked,if we had gone with the “−” then we could not still

satisfy the initial conditions, i.e., y = −2−√12(x2 + 1)

has y(√7) = −4 6= 0.)

Quiz 13 problem bank

1. Find limθ→0

3 sin(θ) − sin(3θ)θ3

.

2. Find limx→0

cos(3x) − 1− x2

ex + e−x − 2.

3. Find limt→∞ t

137

et.

4. Find limx→∞

(1− e−x

)x.

5. Find limx→0

(cos x

)1/x2 .

6. Now, the Star-Belly Sneetches had bellies withstars. The Plain-Belly Sneetches had none uponthars. Then one day Sylvester McMonkeyMcBean came to town with his wondrouslywonderful machine, “Just one pass through,hop on board, and you will have a star forsure.” The Sneetches listened and the Sneetchesthought and those who wanted a star-bellystepped up and bought. Sylvester kept track ofthe proportion of Sneetches with stars (P) andnoticed with time (t) in months thatP ′ = 1

3t(1− P)2. His business was quick, he did

not want to delay, and so he recalled on his veryfirst day that P = 1/4, to make a quick buck andthen leave this place he decided to leave whenP = 3/4. How many months then will it takeuntil Sylvester McMonkey McBean leaves thisplace?

7. It has recently been revealed that the reasonthat twinkies have such an incredible shelf lifeis due to a rare molecule known as Twinkonium(T). Each twinkie contains five grams ofTwinkonium when first produced,unfortunately the Twinkonium then starts todecay and turn into ordinary sugar, throughextensive research it has been determined thatthe rate of decay satisfies T ′ = − 1

10T2, where t

is time measured in months. If a twinkie is stillgood when it has at least one gram ofTwinkonium, how many months from when itwas first produced will the twinkie go bad?

8. Find y given y ′ = 2x(y2 + 1) and y(2) = 0

9. Find y given y ′ =x

e2y + eyand y(

√3) = 0.

10. [Mystery problem]