7.1 Inverse Functions and their Derivatives

2 groups of functions:1. Algebraic- (ex. polynomial) functions obtained by addition, subtraction, multiplication, division, or by taking powers and roots.

2. Transcendental – not algebraic (ex. trigonometric, exponential, logarithmic)

Inverse of f = a function that undoes or inverts

** a function is a rule that assigns a value from its range to each element in its domain some functions assign the same range value to more than 1 element in the

domain Ex. f(x) = x² assigns the same value 1 to -1 and 1 other functions assume each value in their range no more than once …

One to One function = a function f(x) is one to one on a domain, D, iff(x1) ≠ f(x2) whenever x1 ≠ x2 in D*functions take on any one value in their range exactly once

Horizontal Line Test for 1 to 1 Functions: a function y=f(x) is 1:1 if and only if its graph intersects each horizontal line at most once

Inverse Functions - suppose that f is 1:1 functions on its domain D, with range R. ThenThe inverse f-1 is defined by: f-1(a) = b if f(b) = a The domain of f-1 = range of f(x) The range of f-1 = domain of f(x)If you compose a function with its inverse (f-1◦f)(x) = x only a 1:1 function can have an inverse

Functions with Inverses:1. function that is increasing on an interval is 1:1 (makes f(x2)>f(x1), when x2>x1)2. Decreasing functions3. Functions that have positive derivatives at all x are increasing so have inverses4. Functions with negative derivatives at all x are decreasing so have inverses5. Functions that are neither decreasing nor increasing may still be 1:1 and have an

inverse (ex. sec-1x)

Process of passing from f to f -1 can be summarized as a 2- step process: #1. Solve the equation y = f(x) for x this gives a formula x=f-1(y) whose x is expressed as a function of y.

#2. Interchange x and y, obtaining a formula y=f-1(x) where f-1 is expressed in the conventional format with x as the independent variable and y is the dependent variable

Derivatives of Inverses of Differentiable Functions

*Derivatives of f(x) and f-1(x) will be reciprocals of each other*(when graphed – slopes are reciprocals of each other at corresponding points)

Theorem 1: The Derivative Rule for Inverses*If f has an interval I as domain and f-1(x) exists and is never 0 on I, then f-1 is differentiable at every point in its domain. The value of (f-1)′ at a point, b, in the domain of f-1 is the reciprocal of the value of f′ at the point a = f-1(b)

(f-1)′(b) = 1__ or df -1 | 1 f′(f-1(b)) dx |x=b df |

dx|x=f-1(b)

**What says: --when y=f(x) is differentiable at x=a, and we change x by a small amount dx, the corresponding change in y is approximately dy=f′(a) dx--means that y changes @ f′(a) times as fast as x when x=a and that x changes @ 1/f′(a) times as fast as y when y=b. --It is reasonable that the derivative of f-1 at b= the reciprocal of the derivative of f at a.**The derivative of f evaluated at a point and the derivative of the inverse evaluated at the point at which f was originally was evaluated are derivatives:

Let f(x) = x³ -2. Find the value of df-1/dx at x = 6 = f(2)Df/dx = 3x²|x=2 = 12

Now: df-1/dx = 1 = 1Df/dx|x=2 12

Ex’s..

7.2 Natural Logarithms

Exponential Form: f(x) = ax – easy to define when x is an integer/rational #--when irrational the meaning is not as clear

Now we will study the Natural Log function defined as an integral through the Fundamental Theorem of Calculus.

The Natural Logarithm Function As an Integral:x

ln x = ∫ 1/t dt x>0 1

a) if x>1 then ln x is the area under the curve y = 1/t from t=1 to t=xb) for 0<x<1, ln x gives the negative of the area under the curve from x to 1c) for x≤0 function is not defined

1

Zero Width Interval Rule ln 1 = ∫ 1/t dt = 0 1

The Number e = that # in the domain of the natural logarithm satisfying ln e = 1**Geometrically – corresponds to: the point on the x-axis for which the area under the graph of y=1/t and above the interval [1,e] is the exact area of the unit square

Derivative of y = ln x for every positive value of x, we have:

d ln x = 1 dx x

** derivative is always + so the natural logarithm is an increasing function Therefore, it is 1:1 and invertible

** If u is a differentiable function of x whose values are +, so that ln u is defined then apply the chain rule:

d ln u = 1 du if u>0 dx u dx

Ex. d ln 2x = 1 (2) = 1 dx 2x x

y=ln 2x has the same derivative as y = lnx this is true of y = ln ax for any positive #a because constant ends up cancelling out.

Ex. d ln 4x = 1 (4) = 1 dx 4x x

Properties of Logs – for any #’s a>0 and x>0, the natural log satisfies the following:1. Product Rule: ln ax = lna + lnx2. Quotient Rule: ln a/x = lna – lnx3. Reciprocal Rule: ln 1/x = -lnx4. Power Rule: lnxr = rlnx

Ex. ln 10 = ln(5∙2) = ln 5 + ln2Ex. ln2 – ln3 =Ex. ln1/5 =

Applying the propertiesEx. ln 6 + ln cos x =Ex. ln (5x+2) =

(3x-1)Ex. ln 5√x+2

The graph and range of ln xa) the 1st derivative of ln x = 1/x is positive for x>0 and is increasingb) the 2nd derivative of ln x = -1/x² so graph is concave down

lim ln x = ∞ and lim ln x = -∞x∞ x0+

Domain: set of all positive real #’s Range = entire real line giving the graph of lnx

The Integral ∫1/u du

** If u is a differentiable function that is never 0 ∫1/u du = ln|u| + C

Ex.:

** use above formula only if matches that form. – if not do normal integration

Integrals of tan x and cot x

∫tan x dx = ∫sinx dx = ∫-du cosx u

=-∫du = -ln|u| + c u= -ln|cos x| + C ln 1 + C

|cos x|= ln |sec x | + C

Or:

∫tan u du = -ln|cos u| + c = ln|sec u| + c

∫cot u du = ln|sin u| + c = -ln|csc u| + c

Ex.

Logarithmic Differentiation = Using the law of log to simplify the formulas before differentiating: (think back to implicit differentiation)

Ex. find dy/dx of y = (x²+1)(x+3) 1/2 x>1 x - 1

steps:

HI!!!

7.3 Exponential function x=ex (the inverse of ln x)

**Remember ln x is an increasing function of x with domain (0,∞) with range (-∞,∞) it has an inverse (ln-1) x with domain (-∞,∞) with range (0,∞)

domain: lim ln-1x = ∞ range: lim ln-1x = 0 x∞ x-∞

ln e = 1 so e = ln-1(1) e = 2.718281828459045…

y = ex e² = e∙e e-2 = 1/e² e1/2=√e

**er has a logarithm ln er = rlne = r∙1 = r

The natural Exponential Function = for every real # x, ex = ln-1x = exp x

Inverse Equations for e x and ln x a) eln x = x (all x>0)b) ln(ex) = x (all x)

Domain of ex = (-∞, ∞)Range of ex = (0, ∞)

Ex.

General Exponential Functions – for any #’s a>0 and x, the exponential function with base a is ax = exlna

ex. 2√3

2π

Laws of Exponents for ex

1. ex1∙ex

2 = ex1+x2

2. e-x = 1/ex

3. ex1/ex

2 = ex1-x2

4. (ex1)x

2 = ex1x2 = (ex2)x1

Derivative and Integral of e x = exponential function is differentiable b/c it is the inverse of differentiable function whose derivative is never = 0

*The natural exponential function ex is its own derivative d ex = ex

dx

Chain Rule of ex = d eu = eu du dx dx

Ex…

The number e expressed as a limit – can be calculated as the limit e = lim (1+x)1/x

x0

Power Rule (general form) – if u is a positive differentiable function of x and n is any real #, then un is a differentiable function of x and

d un = nun-1 du dx dx

Ex.

Integral of ex = ∫eu du = eu + c

Ex.

Initial Value Problems:

Ex. ey dy/dx = 2x x>√3 y(2)=0

Steps:

7.4 ax and logax

The derivative of a u ** If a>0 and u is a differentiable function of x, then au is a differentiable function of x and:

d au = auln a du dx dx

ex. d/dx 3x = 3xln 3

ex. d/dx 3sinx

The derivative of ax is positive if a>1, and negative if 0<a<1**increasing if >1 and decreasing if 0<a<1** ax is 1 to 1** 2nd derivative d²/dx² (ax) = d/dx (axln a) = (ln a)²ax

is positive of all x so the graph of ax is concave up on each interval

The Integral of a u -- if a ≠ 1 so that ln a ≠ 0 then∫au du = a u + c ln a

ex.

Logarithms with Base a**If a is any positive # other than 1, the function ax is one to one and has a nonzero derivative at every point therefore, has a differentiable inverse

Inverse is logarithm of x with base a denoted by logax

logax is the inverse function of ax (graph is the reflection of ax across line y=x)

Inverse Equations for a x and log ax 1. alog

ax = x for x >0 ex. 2log

23 = 3

2. loga(ax) = x ex. log2(25) = 5

Change of Base Formula:*logax = 1 ∙ ln x = ln x ex. log102 = ln2 ln a ln a 1n10

** All properties also apply to logarithms product/power/quotient

Derivative of logax -- must convert to natural logs to find

d/dx (logau) = d ln u = 1 ∙ 1 du dx ln a ln a u dx

Ex. d/dx log10(3x+1)

Ex. Log25ex – log5√x

Integral of logax - convert to change of base: ln x then 1 ∫ u du Ln a ln a

ex. ∫log2x dxx

7.5 Exponential Growth and Decay

**Increases or decreases in a function with changes in the independent variable

**In real world situations a quantity y, increases/decreases at a rate proportional to size at a given time t.

Law of Exponential Change: y = y0ekt y0 = initial condition k>0 growth k<0 decay k=rate constant of equation

**says that the rate of change is proportional to what has already been accumulated

**if we derived the above it shows that the only factors that are their own derivatives are constant multiples of the exponential function

Compounding Interest Continuously = A(t) = A0ert A0 = initial amountt=years r = continuous interest rate

Half-Life = the time required for half-life of the radioactive nuclei present in a sample to Decay.** a constant – that doesn’t depend on the # of radioactive nuclei initially present

in the sample, but only on the radioactive substance

Half-life = ln2 k

7.7 Inverse Trig. Functions

Inverse Trig Functions:1. are not 1:1 their values repeat periodically so therefore do not have an inverse2. However, can restrict their domain to intervals on which they are 1:1

Original Function Domain Range Sin x [-π/2,π/2] [-1,1]

Cos x [0,π] [-1,1]

Tan x (-π/2,π/2) (-∞,∞) has asymptotes

Cot x (0,π) (-∞,∞)

Sec x [0,π/2)U(π/2,π] (-∞,-1]U[1,∞)

Csc x [-π/2,0)U(0,π/2] (-∞,-1]U[1,∞)

Inverses are denoted by:

Y = sin-1x or y= arcsin x

Y= cos-1x or y= arccos x

Y = tan-1x or y = arctan x

Y = cot-1x or y = arccot x

Y= sec-1x or y = arcsec x

Y = csc-1x or y = arccsc x

** -1 in notation means inverse NOT RECIPROCAL

#1. Arcsin x= the angle in the restricted domain: [-π/2,π/2] whose sine is x

a) Sin-1x domain: [-1,1] and range is [-π/2,π/2]

b) arcsin is symmetric about the origin so odd function

c) values come from Q1 and Q4 b/c of specified domain (Q4 is the – radians of Q1)

ex. sin-1½ (saying: “what radian value in my interval gives me a sine of ½)

ex. sin -1(-√2/2)

ex. sin-1(-1)

ex. sin-1(sin π/6)

#2 Arccos x = the angle in the restricted domain [0.π] whose cosine is x

a) cos-1x domain: [-1,1] and range is [0,π]

b) there is no symmetry

c) Values come from Q1 and Q2 b/c of specified domain

Ex. cos-1½

Ex. cos-10

Ex. cos-1(-√2/2)

Ex. cos-1(cos 5π/4)

#3. Arctan x = the angle in the restricted domain (-π/2, π/2) whose tan is x

a) we use open intervals ( ) b/c cotangent and tangent have asymptotes

b) tan-1x domain: (-∞,∞) and range is: (-π/2,π/2)

c) graph of tan is symmetric about origin and odd

d) Values come from Q1 and Q4 – Q4 values are the – radians in Q1

Ex. tan-11

Ex. tan-1 (-√3)

#4 Arccot x = the angle in the restricted domain (0,π) whose cot is x

a) no symmetry in this graph

b) cot-1 x domain: (-∞,∞) and range: (0,π)

c) values are in Q1 and Q2

ex. cot-1(-1)

ex. cot-1(√3)

ex. cot-1(-1/√3)

#5. Arcsec x = the angle in the restricted domain [0,π/2)U(π/2,π] whose sec is x

a) there is no general agreement about how to define sec-1x for negative values of x, so we agree to use Q2 because it makes cos-11/x = sec-1x. It also makes sec-

1x an increasing function

b) sec-1x domain: |x| ≥ 1 and range: [0,π/2)U(π/2,π]

c) Values come from Q1 and Q2

ex. sec-1√2

ex. sec-1(-2/√3)

#6 Arccsc x = the angle in the restricted domain[-π/2,0)U(0,π/2]

a) 0 is an asymptote

b) Csc-1x domain: |x|≥ 1 and range: [-π/2,0)U(0,π/2]

c) Values come from Q1 and Q4 – Q4 are the – radians of Q1

Ex. csc-1(-√2)

Ex. csc-1(2/√3)

In situations when the inverses do not match radian values on the unit circle, we rely on the Pythagorean Theorem to solve:

Ex. find the cos x, tanx, sec x, csc x, cot x, if x = sin-12/3

** remember sin = o/h so draw a right triangle and go from there

Derivatives of Inverse Trigonometric Functions:

#1. Derivative of y = sin-1x

Differentiable on interval [-π/2,π/2]

Its derivative –cosine is positive on this interval

So it applies to theorem 1 in sect. 7.1 will have an inverse so is differentiable on ---1<x<1

Not differentiable at -1 and 1 b/c tangents to the graph are vertical at these points

d/dx (sin-1u) = 1 du/dx |u| < 1

√ 1- u²

Ex. d/dx (sin-1x²) = 1 ∙ 2x = 2x

√1- (x²)² √1-x4

Derivative of y = tan-1u

d/dx (tan-1u) = 1 du/dx

1 + u²

Ex. tan-1√t = 1 ∙ ½t-1/2

1 + (√t)²

Derivative of sec-1u – derivative is positive for 0<x<π/2 and π/2<x<π so is differentiable by theorem 7.1 (having an inverse of a function)

d/dx (sec-1u) = 1 ∙ du |u|√u²-1 dx |u|>1

ex. d/dx sec-1(5x4) = 1 ∙ (20x³) = 4 |5x4|√(5x4)² - 1 x√25x8 -1

Other derivatives: use Inverse Functions Inverse Cofunction Identities:

Cos-1x = π/2 – sin-1x

Cot-1x = π/2 – tan-1x

Csc-1x = π/2 – sec-1x

** the derivatives of the inverse cofunctions are the negatives of the derivatives of the corresponding inverse functions **

d/dx cos-1u = - 1 ∙ du √1 - u² dx

d/dx cot-1u = - 1 ∙ du1 + u² dx

d/dx csc-1u = - 1 ∙ du |u|√u²-1 dx

Integration Formulas: - 3 useful formulas – hold for any constant a ≠0

#1. ∫ du u √a²-u² = sin-1 a + C (valid for u²<a²)

#2. ∫ du 1 ua²+ u² = a tan-1 a + C (valid for all u)

#3. ∫ du 1 |u |u√u²-a² = a sec-1| a | + C (valid for |u| > a > 0)