Calculus - A Basic Rules

Introduction to differentiation

Introduction

This leaflet provides a rough and ready introduction to differentiation. This is a technique usedto calculate the gradient, or slope, of a graph at different points.

The gradient function

Given a function, for example, y = x2, it is possible to derive a formula for the gradient of its graph.We can think of this formula as the gradient function, precisely because it tells us the gradientof the graph. For example,

when y = x2 the gradient function is 2x

So, the gradient of the graph of y = x2 at any point is twice the x value there. To understand howthis formula is actually found you would need to refer to a textbook on calculus. The importantpoint is that using this formula we can calculate the gradient of y = x2 at different points on thegraph. For example,

when x = 3, the gradient is 2× 3 = 6.

when x = −2, the gradient is 2× (−2) = −4.

How do we interpret these numbers? A gradient of 6 means that values of y are increasing at therate of 6 units for every 1 unit increase in x. A gradient of −4 means that values of y are decreasingat a rate of 4 units for every 1 unit increase in x.

Note that when x = 0, the gradient is 2× 0 = 0.

Below is a graph of the function y = x2. Study the graph and you will note that when x = 3the graph has a positive gradient. When x = −2 the graph has a negative gradient. When x = 0the gradient of the graph is zero. Note how these properties of the graph can be predicted fromknowledge of the gradient function, 2x.

When x = 3 the gradient is positiveand equal to 6

When x = −2 the gradient is negativeand equal to −4.

When x = 0 the gradient is zero.x

y

−4−3−2−1 0 1 2 3 4

5

10

15

ExampleWhen y = x3, its gradient function is 3x2. Calculate the gradient of the graph of y = x3 when a)x = 2, b) x = −1, c) x = 0.

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VB234 Apply Calculus to Engineering Problems PART 1. DIFFERENTIATION

Solutiona) when x = 2 the gradient function is 3(2)2 = 12.

b) when x = −1 the gradient function is 3(−1)2 = 3.

c) when x = 0 the gradient function is 3(0)2 = 0.

Notation for the gradient function

You will need to use a notation for the gradient function which is in widespread use.

If y is a function of x, that is y = f(x), we write its gradient function asdy

dx.

dy

dx, pronounced ‘dee y by dee x’, is not a fraction even though it might look like one! This notation

can be confusing. Think ofdy

dxas the ‘symbol’ for the gradient function of y = f(x). The process

of findingdy

dxis called differentiation with respect to x.

ExampleFor any value of n, the gradient function of xn is nxn−1. We write:

if y = xn, thendy

dx= nxn−1

You have seen specific cases of this result earlier on. For example, if y = x3,dy

dx= 3x2.

More notation and terminology

When y = f(x) alternative ways of writing the gradient function,dy

dx, are y′, pronounced ‘y dash’,

ordf

dx, or f ′, pronounced ‘f dash’. In practice you do not need to remember the formulas for the

gradient functions of all the common functions. Engineers usually refer to a table known as a Tableof Derivatives. A derivative is another name for a gradient function. Such a table is available onthe leaflet Table of Derivatives. The derivative is also known as the rate of change of a function.

Exercises1. Given that when y = x2, dy

dx= 2x, find the gradient of y = x2 when x = 7.

2. Given that when y = xn, dydx

= nxn−1, find the gradient of y = x4 when a) x = 2, b) x = −1.

3. Find the rate of change of y = x3 when a) x = −2, b) x = 6.

4. Given that when y = 7x2 + 5x, dydx

= 14x + 5, find the gradient of y = 7x2 + 5x when x = 2.

Answers1. 14. 2. a) 32, b) −4. 3. a) 12, b) 108. 4. 33.

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1 Introduction to differential calculus Differential calculus is the branch of mathematics that deals with the process of differentiation, that is finding gradients of curves, rates of change and applications of differentiation.

2 Limits i) Definition A limit in the mathematical sense is concerned with finding the limiting value of a function ( )f x as x approaches the value of a. The notation x a→ is used to denote as x approaches the value of a The notation lim ( )f x is used to denote the limiting value of the function ( )f x Together lim ( )

x af x

→means the limiting value of the function ( )f x as x approaches the

value of a. Example 1 Find li 2

3m

xx

→

Solution 2

3lim 9x

x→

=

Notice that this does not mean that when 23 then 9x x= =2

, what it states is that as x gets closer and closer to the value of 3 then x gets closer and closer to the value of 9. ii) Theorems on limits

• the limit of a constant function is c limx a

c c→

=

• If lim ( )

x af x

→= L and lim ( )

x ag x M

→= then ( )m ( ) (

x ali f x g x L M

→± = ±

That is the limit of a sum ( or difference ) is the sum ( or difference ) of the limits.

• If lim ( )x a

f x→

= L and lim ( )x a

g x M→

= then ( )m ( ) ( )x ali f x g x LM

→=

That is the limit of a product is the product of the limits.

• If lim ( )x a

f x→

= L and lim ( )x a

g x M→

= then lim ( )( )m

( ) lim ( )x a

x ax a

f xli f x L

g x g x M→

→→

= =

That is the limit of a quotient is the quotient of the limits provided that 0M ≠

1

A. BASIC RULES

Example 2 Find ( )2

2m 2 3 5

xx x

→li − +

Solution

( )2

2

2

2 2 2

lim 2 3 5 by the properties of the limits

lim 2 lim3 lim5

8 6 57

x

x x x

x x

x x→

→ → →

− +

= − +

= − +=

Example 3 Find 2

3

9m3x

xx→

li −−

Solution

( )( )

( )

2

3

3

3

9lim33 3

lim3

lim 3 since 3

6

x

x

x

xxx x

xx x

→

→

→

−=

−+ −

=−

= +

=

≠

Note that we can’t merely substitute in 3x = as this would give 00

which is undefined.

To avoid having 00

or in fact a 0 in the denominator we need to factorize and simplify

before taking the limit.

Example 4 Find 2

22

10 24m4 12x

x xx x→

li + −+ −

Solution

( )( )( )( )

2

22

2

2

10 24lim4 12

2 12lim

2 612lim since 26

14 78 4

x

x

x

x xx xx xx x

x xx

→

→

→

+ −=

+ −− +

=− +

+= ≠

+

=

2

Example 5 Sketch the graph of the function 2 9( )

3xf xx

−=

−

Solution

( )( )2 3 39( )3 3

3 if 3undefined if 3

x xxf xx x

x xx

+ −−= =

− −+ ≠

= =

The function has a domain of { }\ 3R and (3)f is undefined Consider the table of values for ( ) near 3f x x =

x 2.9 2.99 2.999 3 3.001 3.01 3.1 ( )f x 5.9 5.99 5.999 ? 6.001 6.01 6.1

we can see from the table of values that as x approaches the value of 3 from below 3 and from above 3 the limiting value of the function approaches the value of 6. Hence 6 is the limiting value. The function however is undefined at 3x = and the graph below appears to be a straight line with a hole in the graph at the point where 3x = . Such graphs are called discontinous at a point.

We say that as x approaches the value of 3 from ( the left ) or below then 3x −→2

3

9lim 63x

xx−→

−=

− and as x approaches the value of 3 from ( the right ) or above 3x +→

then 2

3

9lim 63x

xx+→

−=

− although both of these limits exist and are equal ( )3f does not exist,

so the function not continous.

3

iii) Infinite limits The symbol infinity ∞ is really just a notation meaning as x gets very very large, and

1lim 0x x→∞

= and 2

1lim 0x x→∞

=

Example 6 Find 3 5m2 5x

xx→∞

li +−

Solution

3 5

2 5

5

5

3 5lim dividing both the numerator and denominator by 2 5

lim

3lim2

32

x

xx

xxx

x

xx

x xx→∞

+

−→∞

→∞

+−

=

+=

−

=

Example 7 Find 2

2

3 5m2 7x

x xx x→∞

45

li + −− +

Solution

2

2

2

2

2

2

22

2

3 5 4

2 7 5

5 4

7 5

3 5 4lim dividing both the numerator and denominator by 2 7 5

lim

3lim

2

32

x

x xx

x xxx

x xx

x x

x x xx x→∞

+ −

− +→∞

→∞

+ −− +

=

+ −=

− +

=

4

3 Gradients of curves The gradient of a curve is defined to be the gradient of the tangent to the curve at that point. The gradient of a curve continually changes at each point as we move along a curve. Example 8 Draw tangents to the curve 2 5 at 3 , 1 ,1 , 3y x x= + = − − Solution

The gradient of a straight line is constant, and is independent of where we are on the line. The gradient of a line is denoted by m and is the same at all points on the line. Parallel lines all have the same gradient and lines sloping upwards to the right have a positive gradient, while lines sloping upwards to the left have a negative gradient. 0m0m > <

5

4 First principles Consider the curve we want to find a formulae which gives the gradient of the curve at a general point P with coordinates

( )y f x=( )y f x= ( ), ( )P x f x on the curve.

To do this consider a “neighbouring” point Q on the curve ( )y f x= , by this we mean that the point Q is close to the point P in fact it is a small horizontal distance h away, so that the coordinates of Q are ( ), ( )h f x h+ +Q x

Now the gradient of the line segment joining the points P and Q is given by

( ), ( )P x f x

( , (Q x h f

( )f x h+ − (f x)

h

y

x

))x h+ +

( ) (( ) )f x h f xm PQ

h+ −

=

Imagine now as the points P and Q get closer and closer to each other. That is the horizontal distance between them h approaches zero. In the limit as h approaches zero, the points P and Q coincide and we have the gradient at P. The gradient at this point is denoted by ( )f x′

0

( ) (( ) limh

)f x h f xf xh→

+ −′ =

Using this formula to obtain the gradient function is called using the METHOD OF FIRST PRINCIPLES. The actual process of finding the gradient function from the original function is called the process of DIFFERENTIATION. There are alternative notations for this and often we use the symbol x∆ ( this is one single quantity ) meaning the small change in x instead of h and in this case the formulae

is written as 0

limx

dy ydx x∆ →

∆=

∆

6

Example 9 For the function 2( ) 4 12f x x x= + − find i) the gradient function using the method of first principles ii) where the tangent to the curve is parallel to the x axis iii) and explain what they represent (3) and (3)f ′fiv) the equation of the tangent to the curve at the point where 3x = v) Sketch the curve showing the tangent at 3x = vi) { }: ( ) 0x f x >

vii) { }: ( ) 0x f x′ > Solution

( )( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( )

( ) ( )

( ) ( )( )

2

2

0

2 2 2

0

2

0

0

0

i) 4 12

4 12

lim

2 4 4 12 4lim

2 4lim

2 4lim

lim 2 4 0

2 4

h

h

h

h

h

f x x x

f x h x h x h

f x h f xf x

hx xh h x h x x

f xh

xh h hf xh

h x hf x

hf x x h h

f x x

→

→

→

→

→

= + −

+ = + + + −

+ −′ =

+ + + + − − + −′ =

+ +′ =

+ +′ =

′ = + + ≠

′ = +

12

ii) For the tangent to the curve to be parallel to the x axis we require that the gradient is zero. This will give the x coordinate of the turning point or the vertex of this parabola

( ){ }( )

( )

we need to solve : 0

2 4 0 so that 2 42

Now when 2 2 4 8 12 16

x f x

f x x xx

x y f

′ =

′ = + = = −

= −

= − = − = − − = −

The point is ( ) { } [ )2, 16 the range of the function is : 16 16,y y− − ≥ − = − ∞

iii) ( ) ( )3 9 12 12 9 and 3 6 4 10f f ′= + − = = + = When 3 the coordinate is 9 and the gradient at this point is 10x y=iv) to find the equation of the tangent to the curve at the point ( )3,9 where the gradient

is m we use the result 10T = ( )1 1Ty y m x x− = − so that

7

( )9 10 3 or 10 21y x y x− = − = −

v) to sketch the graph ( ) ( )( )2 4 12 6 2y f x x x x x= = + − = + −

the graph crosses the x axis at 6 and 2x x= − = as coordinates ( ) (6,0 and 2,0− )

the graph crosses the y axis at 12y = − as a coordinate ( )0, 12 −

vi) { }: ( ) 0x f x > This means find the x values for which the function is positive { } { } ( ) ( ): 2 : 6 2, , 6x x x x> ∪ < − = ∞ ∪ −∞ − vii) { }: ( ) 0x f x′ > This means find the x values for which the gradient is positive { } ( ): 2 2,x x > − = − ∞

8

5 Basic rules for differentiation i) Derivative of ( )f x x=

0

0

0

0

( )( )

( ) (( ) lim

lim

lim

lim 1

1

h

h

h

h

f x xf x h x h

)f x h f xf xh

x h xh

hh

→

→

→

→

=+ = +

+ −′ =

+ −=

=

=

=

ii) Derivative of 2( )f x x=

( )

( )

( )

2

2 2 2

0

2 2

0

2

0

0

0

( )

( ) 2( ) ( )( ) lim

2lim

2lim

2lim

lim 2 0

2

h

h

h

h

h

f x x

2

f x h x h x xh hf x h f xf x

hx xh h x

hxh h

hh x h

hx h h

x

→

→

→

→

→

=

+ = + = + +

+ −′ =

+ + −=

+=

+=

= + ≠

=

9

iii) Derivative of ( ) 3f x x=

( )

( )

( )

3

3 3 2 2

0

3 2 2 3 3

0

2 2 3

0

2 2

0

2 2

0

2

( )

( ) 3 3( ) ( )( ) lim

3 3lim

3 3lim

3 3lim

lim 3 3 0

3

h

h

h

h

h

f x x3f x h x h x x h xh h

f x h f xf xh

x x h xh h xh

x h xh hh

h x xh hh

x xh h h

x

→

→

→

→

→

=

+ = + = + + +

+ −′ =

+ + + −=

+ +=

+ +=

= + + ≠

=

iv) Derivative of 4( )f x x=

( )

( )

( )

4

4 4 3 2 2 3

0

4 3 2 2 3 4 4

0

3 2 2 3 4

0

3 2 2 3

0

3 2 2 3

0

3

( )

( ) 4 6 4( ) ( )( ) lim

4 6 4lim

4 6 4lim

4 6 4lim

lim 4 6 4 0

4

h

h

h

h

h

f x x4f x h x h x x h x h xh h

f x h f xf xh

x x h x h xh h xh

x h x h xh hh

h x x h xh hh

x x h xh h h

x

→

→

→

→

→

=

+ = + = + + + +

+ −′ =

+ + + + −=

+ + +=

+ + +=

= + + + ≠

=

10

v) Derivative of 1( )f xx

=

( )( )

( )

0

0

0

0

2

1( )

1( )

( ) (( ) lim

1 1

lim

lim

1lim 0

1

h

h

h

h

f xx

f x hx h

)f x h f xf xh

x h xh

x h xhx x h

hx x h

x

→

→

→

→

=

+ =+

+ −′ =

−+=

+ −=

+

= ≠+

= −

vi) Derivative of ( )f x x=

( )( )

0

0

0

0

0

( )

( )( ) ( )( ) lim

lim

lim x

lim

1lim 0

12

h

h

h

h

h

f x x

f x h x hf x h f xf x

hx h x

hx h x x h

hx

x h xx h x

h x h x

hx h x

x

→

→

→

→

→

=

+ = ++ −′ =

+ −=

+ − + +=

+ ++ −

=+ +

= ≠+ +

=

11

vii) Summary From the above results we have the following table

y

dydx

4x 34x 3x 23x 2x 2x

x 1 11

x x−= 221

xx−− = −

viii) Rule 1: Power rule The procedure to differentiate a power of x, the result is the power times x

to the one less power. To differentiate a power of x that is ( ) 1n nd x nxdx

−=

This rule is valid for all values of n Example 10

Find dydx

if i) ii) 101y x= 2

1yx

= iii) y x=

Solution

2

2

33

1ii)

2 1 322

y xx

n ndy xdx x

−

−

= =

= − − = −

= − = −

12

12

12

1 12 2

12

iii)1

12

y x xn n

dy xdx x

−

= == − = −

= =

101

100

i)101 1 100

101

y xn n

dy xdx

== − =

=

ix) Rule 2: Derivative of constants Since any line parallel to the x axis has a zero gradient it follows that if C is a constant

then 0

the derivative of a constant is zero

dyy Cdx

= =

x) Rule 3: Term by term differentiation We can differentiate term by term so long as the functions are added or subtracted. If are both function of x then if ( ) and ( )u u x v v x= =

then dy du dvy u vdx dx dx

= ± = ±

These rules can be proved from first principles and the properties of limits.

Example 11 If then find 5 3y x x x= − +dydx

Solution

5 3

4 25 3

y x x xdy x xdx

= − +

= − +1

xi) Rule 4: Constant multiples Constants which multiply the function merely multiply the derivatives, that is if

1 is a constant and then n ndya y ax naxdx

−= =

Using these basic rules we can merely write down the derivative of polynomial functions Example 12 If ( )2( ) 3 6 5 find f x x x f x′= − + Solution

( )

2( ) 3 6 5 then 6 6f x x xf x x

= − +′ = −

In many other differentiation problems it may be first necessary to simplify the function, before we perform the differentiation process.

13

Example 13 Differentiate ( )223 5x − Solution

( ) ( )2 22 2

4 2

3

let 3 5 expanding first using 2

9 30 25 now differentiating

36 60

y x a b a ab b

y x xdy x xdx

= − − = − +

= − +

= −

2

Example 14 Differentiate 23 52

xx+

Solution 2

2

1

2

2

2

2

3 5let 2

3 5 simplifying2 23 52 23 5 simplifying2 23 52 23 5

2

xyx

xyx xxy x

dy xdxdydx xdy xdx x

−

−

+=

= +

= +

= −

= −

−=

Example 15 Differentiate 2

3 52xx

−

Solution

12

32

52

2

2 2

2

3

35

3

3 5let 2

3 5 simplifying using index laws2 23 52 2

9 54

9 54

20 94

xyx

xyx x

y x x

dy x xdxdydx xxdy xdx x

− −

− −

−=

= −

= −

= − +

= − +

−=

Exercise 1.

14

6 The chain rule

The Chain rule states that dy dy dudx du dx

=

this can be proved from first principles and the limit theorems. Example 16 Differentiate ( with respect to x )35 4x +

Solution

( )

( )

3

3

2

2

2

Method II

5 4

where 5 4

3 and 5

15 but 5 4

15 5 4

y x

y u u xdy duudu dxdy dy du u u xdx du dxdy xdx

= +

= = +

= =

= = =

= +

+

( )( )

( )

( )

3

3 3 2 2 3

3 2

2

2

2

Method I

5 4

expanding using 3 3

125 300 240 64 differentiating gives

375 600 240

15 25 40 16

15 5 4

y x

a b a a b ab b

y x x xdy x xdxdy x xdxdy xdx

= +

+ = + + +

= + + +

= + +

= + +

= +

Example 17 Given that 1( ) find (1)4 5

f x fx

′=+

Solution

( )

( )

1

1

2

22

2

1 4 54 5

where 4 5

and 4

44

4( )4 5

4(1)81

y xx

y u u xdy duudu dxdy dy du udx du dx u

dyf xdx x

f

−

−

−

−

= = ++

= =

= − =

= = − = −

′ = = −+

′ = −

+

15

Example 18 Determine 9 4d xdx

+

Solution

( )12

12

121

2

let 9 4 9 4

where 9 41 and 9

29

29

2 9 4

y x x

y u u xdy duudu dxudy dy dudx du dx udydx x

−

= + = +

= = +

= = =

= =

=+

In general if ( ) ( ) 1 then n ndyy ax b na ax bdx

−= + = +

Example 19 Differentiate 2

14 9x +

with respect to x

Solution

( )

( )

122

1 2

22

2

22

1let 4 94 9

where 4 91 and 8

8

8

4 9

y xx

y u u xdy duu xdu u dxdy dy du xdx du dx udy xdx x

−

−

−

= = ++

= = +

= − = − =

= = −

−=

+

16

Example 20 Given that ( )22

1( ) find ( )5

f x fx

x′=+

Solution

( )( )

( )

( )

2222

2 2

3

33

32

1let 55

where 5

2 and 2

44

4( )5

y f x xx

y u u xdy duu xdu dxdy dy du xxudx du dx u

dy xf xdx x

−

−

−

−

= = = ++

= = +

= − =

= = − = −

′ = = −+

Example 21 If ( ) 24x= + 9f x find the coordinates on the curve where the tangent is parallel to the line 3y x= +

Solution

( )

( ) ( )

12

12

12

2 2

2

12

2

2

2 2 2

2

let 4 9 4 9

where 4 92 and 8

4 we require such that 14 9

so that 4 4 9 squaring both sides16 4 9 so that 12 9

3 3 or the co4 2

y x x

y u u xdy duu xdu dxu

dy xf x x f xdx x

x xx x x

x x

−

= + = +

= = +

= = =

′ ′= = =+

= +

= + =

= = ±3 3ordinates are , 2 3 and , 2 3

2 2

−

In general if ( ) ( ) 1 then ( )

n ndyy f x n f x f xdx

−′= =

Example 22 Determine ( )43 22 5d x x xdx

− +

Solution

( )

( )(

43 2

32 3

2 5

4 6 10 1 2 5

d x x xdx

)2x x x x

− +

= − + − + x

17

7 The product rule The product rule is used to differentiate products of two types of functions. If then where and are both functions of , that is ( ) and ( )y uv u v x u u x v v x= = =dy dv duu vdx dx dx

= + this rule can be proved from first principles and the limit theorems.

Example 23 Differentiate ( )( )2 23 5 4 3x xy x= + + − Solution

( )( )

( )( ) ( )

2 2

2 2

2 2

3 2 3 2

3 2

Method I

3 5 4 3

where 3 5 and 4 3

Now 6 and 2 4

3 5 2 4 6 4 3 expanding

6 12 10 20 6 24 18

12 36 8 20

y x x x

y uv u x v x xdu dvx xdx dx

dy dv duu vdx dx dxdy x x x x xdxdy x x x x x xdxdy x x xdx

= + + −

= = + = + −

= = +

= +

= + + + + −

= + + + + + −

= + − +

( )( )2 2

4 3 2 2

4 3 2

2 2

Method II

3 5 4 3 expanding first

3 12 9 5 20 153 12 4 20 15 now differentiating

12 36 8 20

y x x x

y x x x x xy x x x xdy x x xdx

= + + −

= + − + + −

= + − + −

= + − +

Both methods produce the same result, and in this case above the second method is probably the preferred method, however in other cases when we can not simply first before we differentiate, we have no alternative but to use the product rule.

18

Example 24 Find ( )32 if 3 5 4y x xdx

= +dy

Solution

( )( )

( )

( ) ( )

( ) ( )

( )( )

32

32

2

3 22

2

2

3 5 4

where 3 and 5 4

6 and 15 5 4 from the chain rule

6 5 4 45 5 4

3 5 4 2 5 4 15

3 25 8 5 4

y x x

y uv u x v xdu dvx xdx dxdy x x x xdxdy x x x xdxdy x x xdx

= +

= = = +

= = +

= + + +

= + + +

= + +

Example 25 Find 2 if 5 4 9y x xdxdy

= +

Solution

( )

( )( )

( )( )

12

12

2

2

2

22

2

2 2

2

2 2

2 2

5 4 9

where 5 and 4 945 and from the chain rule

4 920 5 4 94 9

20 5 4 9

4 9

40 45 4 9

5 8 9 4 9

y x x

y uv u x v xdu dv xdx dx xdy x xdx x

x xdydx xdy x xdxdy x xdx

−

−

= +

= = = +

= =+

= + ++

+ +=

+

= + +

= + +

19

8 The quotient rule The quotient rule is used to differentiate quotients of two types of functions.

If where and are both functions of , that is ( ) and ( )uy u v x u u x vv

= = v x= then

2

du dvv udy dx dxdx v

−=

Example 26 Find 4 3 if 3 5

dy xydx x

+=

−

Solution

( ) ( )( )

( )

2

2

4 3 3 5

where 4 3 and 3 5

now 4 and 3

4 3 5 3 4 33 5

293 5

x uyx v

u x v xdu dvdx dx

x xdydx xdydx x

+= =

−= + = −

= =

− − +=

−

−=

−

Example 27 Find 2

2

3 if 2 5

dy xydx x

=+

Solution

( )( )

( )

2

2

2 2

2 3

22

22

3 2 5

where 3 and 2 5

now 6 and 4

6 2 5 12

2 5

30

2 5

x uyx v

u x v xdu dvx xdx dxx x xdy

dx x

dy xdx x

= =+= =

= =

+ −=

+

=+

+

20

Example 28 Find 22 5 if 3

dy xydx x

+=

Solution

2

2

1

2

2

Method IIsimplify first

2 5 3

2 53 32 53 32 53 32 53 3

xyx

xyx xxy x

dy xdxdydx x

−

−

+=

= +

= +

= −

= −

( )( )

2

2

2 2

2

2

2

2

2 2

2

Method Iusing the quotient rule

2 53

where 2 +5 and 3

now 4 and 3

12 3 2 5

3

6 159

6 159 92 53 3

x uyx vu x v

du dvxdx dx

x xdydx x

dy xdx xdy xdx x xdydx x

+= =

= =

= =

− +=

−=

= −

= −

x

Example 29 Find 2

3 if 4 9

dy xydx x

=+

Solution

( )

( )32

2

2

2

22

2

2

2 2

2

2

2

3 4 9

where 3 and 4 94now 3 and from the chain rule

4 9123 4 94 9

4 93 4 9 12

4 94 9

27

4 9

x uyvx

u x v xdu dv xdx dx x

xxdy xdx x

x xdy xdx xdydx x

= =+

= = +

= =+

+ −+=

++ −

+=+

=+

Exercise 2

21

9 Derivative of exponential functions i) Derivative of ( ) xf x a=

( )

0

0

0

0

0

( ) using the method of first principles( )

( ) ( )( ) lim

lim

lim

1lim

1lim 0

x

x h

h

x h x

h

x h x

h

x h

h

hx

h

f x af x h a

f x h f xf xh

a ah

a a ah

a ah

aa hh

+

→

+

→

→

→

→

=

+ =+ −′ =

−=

−=

−=

−= ≠

The graphs below show the shape of the graphs of xy a= for various values of a

2 , 3 , and 10a e= and the tangents to the curves at the point where they cross the y axis.

2xy = 3xy =

22

xy e= 10xy =

Curve slope at 0x =

2xy = 0.693m =3xy = 1.099m =

xy e= 1m =10xy = 2.303m =

The Euler number e is defined so that the gradient of the curve xy e= at the point where it crosses the y axis, that is at 0x = is precisely equal to 1. In fact 2 , e is an irrational number and its value e3e< < 2.718≈ This requirement

then means that when a e=0

1im 1 so that t e of is itselfh

x

h

e eh→

−=l he derivativ

That is if then x xdyy e edx

= =

ii) Derivative of y kxe=

where using the chain rule

and

kx u

u

kx

y e e u kxdy due kdu dxdy dy du kedx du dx

= = =

= =

= =

In general if then kx kxdyy e kedx

= = here we a function whose derivative is proportional

to itself. This result has many applications.

23

Example 30 Differentiate each of the following with respect to x

i) ii) 24 xy e= 3

16 xe

=y iii) 636 x=y e

Solution

3

3

3

1ii) 616

12

x

x

x

ye

y e

dy edx

−

−

=

=

= −

2

2

i) 4

8

x

x

y edy edx

=

=

6

3

3

iii) 366

18

x

x

x

y ey e

dy edx

=

=

=

Example 31 Differentiate each of the following with respect to x i) ii) ( 225 3xy e−= + ) ( )23 3x xe−= −y e Solution

( )23 3

6 6

6 6

ii)

2

6 6

x x

x x

x x

y e e

y e edy e edx

−

−

−

= −

= − +

= −

( )22

4 2

4 2

i) 5 3

25 30 9

100 60

x

x x

x x

y e

y e edy e edx

−

− −

− −

= +

= +

= − −

+

Example 32 Differentiate 3 36x x

x

e ee

−+5 with respect to x

Solution

( )3 3

3 3

2 4

2 4

5 6let 5 6

5 6

10 24

x xx x

x

x x

x x

e ey e ee

y e edy e edx

−− −

−

−

+= = +

= +

= −

xe

Example 33 If 324 show that 2 3 0x dy y

dx−= +y e =

Solution 3 32 2

3 32 2

If 4 then 6

so 2 3 12 12 0

x x

x x

dyy e edx

dyLHS y e e RHSdx

− −

− −

= = −

= + = − + = =

24

Example 34 Find 3 2 if 5 xdy y edx

+=

Solution

3 2 3 2

2 3

2 3

3 2

Method IIusing index laws

5 55

15

15

x x

x

x

x

y e e ey e edy e edudy edx

+

+

= =

=

=

=

3 2

3 2

Method Iusing the chain rule

55 where 3 2

5 and 3

15 15

x

u

u

u x

y ey e u xdy duedu dxdy e edx

+

+

=

= =

= =

= =

+

iii) In general

if then kx c kx cdyy Ae kAedx

+ += =

Example 35 Find 3

if xdy y edx

=

Solution

3

3

3

2

2 2

where

and 3

3 3

x

u

u

u x

y ey e u xdy due xdu dxdy x e x edx

=

= =

= =

= =

In general if ( ) ( ) then ( )f x f xdyy e f x edx

′= =

25

Example 36 Find 5 2 if xdy y x edx

−=

Solution

( )

5 2

5 2

4 2

5 2 4 2

4 2

using the product rule

where and

so that 5 and 2

2 5

5 2

x

x

x

x x

x

y x ey uv

u x v edu dvx edx dx

dy dv duu vdx dx dxdy x e x edxdy x e xdx

−

−

−

− −

−

==

= =

= =

= +

= − +

= −

−

Exercise 3

26

10 Derivative of logarithmic functions i) Derivative of y x loge=

log ln

so that

now

1 1

e

y

y

y

y x

x edx edydydx e x

= =

=

=

= =

x

Example 37 find log 3ey = x dy dx

Solution

Method IIUsing log laws

log 3 log 3 log1 10

e ey xdydx x x

= = +

= + =

e x

Method IUsing the chain rule

log 3 log where 31 and 3

3 3 13

e ey x u udy dudu u dxdydx u x x

= = =

= =

= = =

x

Example 38 find 3logey = x dy dx

Solution

3

Method IIUsing log laws

log 3log3

e ey xdydx x

= =

=

3 3

2

2

3

Method IUsing the chain rule

log log where 1 and 3

3 3 3

e ey x u udy du xdu u dxdy xdx u x x

= = =

= =

= = =

x x

27

Example 39 find (log 4 5ey x= )+dy dx

Solution

( )Using the chain rule

log 4 5 log where 4 51 and 4

4 44 5

e ey x u u xdy dudu u dxdydx u x

= + = =

= =

= =+

+

ii) A general result

if ( )log then edy ay ax bdx ax b

= + =+

Example 40 find ( 2log 4 13ey x x= + + ) dy dx

Solution

( )2 2

2

Using the chain rule

log 4 13 log where 4 13

1 and 2 4

2 4 2 44 13

e ey x x u u x x

dy du xdu u dxdy x xdx u x x

= + + = = +

= = +

+ += =

+ +

+

iii) A Final Rule

if ( )log ( ) then ( )e

dy f xy f xdx f x

′= = this is the rule that we will now use, there is no need

to use the chain rule.

28

Example 41 2log 9ey x= + find dy dx

Solution

( )( )

12

2

2

212

2

log 9

log 9 by log laws

log 9

9

e

e

e

y x

y x

y x

dy xdx x

= +

= +

= +

=+

Example 42 3 5log3 5e

xyx

+= −

find dy dx

Solution

( ) ( )( ) (( )( )

)

2

3 5log3 5

log 3 5 log 3 5 by log laws

3 3 5 3 3 53 33 5 3 5 3 5 3 5

309 25

e

e e

xyx

y x x

x xdydx x x x xdydx x

+ = − = + − −

− − += − =

+ − + −

−=

−

29

Exercise 1 - Basic Differentiation 1 Evaluate the following:

(a) li 22

m 3x

x→

(b) 4

m2

lix x→

(c) lix

(d) 3

m (3 4)x→

−2

m2 33 1

xx→

− +

lix

(e) lih

30

m (2 3 )x xh→

− (f) 3

1m

42

xx→

++

lix

(g) 2

5m

255

xx→

−−

lix

(h) 2 2

mlix a

x ax a→

−−

(i) 2

3m

63

x xx→−

+ −+

lix

(j) 2

4m

2 3 24

x xx→

− −−

lix

0

(k) 3

3m

273

xx→

−−

lix

(l) 3 3

mlix a

x ax a→

−−

(m) 2

0m

5 2lih

x h xhh→

− + h (n) 2

23m

5 64 3

x xx x→

− +− +

lix

(o) 2

4m

8 14

x xx→−

+ ++

lix

(p) 6 2

22m

3 22

x xx x→−

+ ++ −

lix

2 Find '( )f x using the definition of the derivative (that is first principles): (a) ( ) 2 2f x x= − x (b) ( ) 2 1f x x x= + + 3 Find the derivative (from first principles) of: (a) f x (b) f x 2( ) 3 5x= − 2( ) 3 21x= + (c) ( ) 4 5f x = + x 3 (d) ( ) 2f x x= −

BASIC RULES - EXERCISES

4 For the function 2( ) 3 5 2f x x= − + x :

(a) find the gradient of the chord joining the points ( ) (, ( ) and , ( )P x f x Q x h f x h+ + ) , (b) deduce the gradient function, and find its value when x = 2 . 5 Find, from first principles, the gradient function of 2( ) 8 16f x x x= − + . Hence, find and state what it represents. '(4)f 6 Find '( )f x for each of the following from first principles:

(a) f x (b) ( 2( ) 2 3x= + ) 1( )f xx

=

7 Find dy from first principles: dx

(a) y x (b) 22 3= − x2

6 42xy x= − +

8 (a) For the graph of (2 1)(3 )y x x= − − , determine dydx

.

(b) If y x , find 2(2 3)= +dydx

.

9 (a) A straight line has the equation y mx c= + . Show from first principles that the gradient of the line is m . (b) The general form of a quadratic function is 2y ax bx c= + + (parabola). Show from

first principles that the gradient function 2dy ax bdx

= + .

10 (a) Find the gradient of the chord PQ to the function if the ( ) ( 2)f x x x= + x-coordinates of P and Q are -1 and 1 h− + respectively.

(b) Hence, find the gradient of the function at P .

11 Let P and Q be the points on the curve at which x = 2 and 22y x= 2x h= + respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve 22 at 2y x x= = . 12 Find the derivatives of each of the following:

(a) 21x

(b) 1x

(c) 3 2x (d) 4 x

(e) 41x

(f) 3

1x

(g) 3

12x

(h) 200x

(i) 1.41

x (j) 0.9

1x

(k) 3

15x

(l) /p qx

13 Find dy for each of the following: dx

(a) y x (b) 22 3x= − + 5 21 8 24

x 1y x= − + −

(c) y a (d) y x2x bx c= + + 3 22 6 6x x 6= − + −

(e) 3 21 1 1013 2

x x= − + + +y x (f) y a 3 2x bx cx d= + + +

(g) y x (h) y x4 3 2 1x x= − − − + 4 3 22 4 2 3 1x x x 8= − + − +

14 (a) Find 2( 3d )x xdx

− (b) Determine 22 x

dx x d x +

15 Differentiate each of the following with respect to x: (a) 2 33 2 5x x− + x ) (b) 3 (4 3x x−

(c) 1xx

+ (d) 22

1xx

+

(e) (3 (f) 24)x −2 3 4x

x− +x

(g) 2

23xx− + 4x (h)

2 3 4xx

− +x

16 For the curve 2( ) (2 3 1)f x x x x= − +

5

6

find an expression for the gradient of the curve at any x value. Find . '(4)f 17 Show that the curve has zero gradient when x = 1 . 22 4y x x= − + 18 Determine the coordinates of the point on the curve 2 4y x x= − + where the gradient is parallel to the x-axis . 19 Find the slopes of the curves at the points indicated: (a) y x at (-2 , 6) (b) 4 3x= + − 4 3( ) 3f x x= + x

7

at x = 1 20 For the curve : 22 8y x x= − +

(a) find the gradient of the curve when x = 2 , (b) What are the coordinates of the point where the tangent to the curve is parallel to the x-axis ?

Exercise 2 - Further Rules of Differentiation 1 Differentiate each of the following with respect to x by applying the chain rule:

(a) 9 5x− (b) 13 4x −

(c) 16 5x−

(d) ( )2

12 3x−

(e) 17 8x +

(f) ( )33 2 1x x x− + −

(g) ( (h) )23 2 1x x x+ + −1

5 7x −

(i) ( )723 15x x− (j) 3

2 1x xx

+ +

(k) 2 8+x (l) 28 x− (m) 2x −8 (n) 7 37 8x− (o) 2 3x+ − 7x (p) 2x x+ −4 3 9 (q) 2 10+ −x x (r) 212 1113 x x+ −

(s) 2

110x +

(t) 2

19 x−

(u) 2

13 2 1x x+ +

(v) 2

17 6x x− + 5

2 Differentiate each of the following with respect to x: (a) 3 x (b) 2 1+ 5 2x x−

(c) ( )47 x− (d) ( ) 621 5x−

+

(e) 71x5

x −

(f) 21

3x+

(g) 17 3x−

(h) ( )51 x+

(i) 22

2 x− (j) 4 53(1 )x− −

(k) 24 1 x+ (l) ( )31 3x+

(m) ( )31 7x+ (n) ( ) 327 3x−

−

(o) 13 x− (p) 3

1(7 8 )x+

(q) ( (r) )5/ 421 3x x− + ( )3/ 221 x−

(s) ( (t) )92 9x −5

317 1x

+

(u) 31 3x x− − (v) ( )329 5 2x x+ +

3 (a) Given that 24y = − x verify that 0dyy xdx

+ = .

(b) If 24y x= 1− show that 21dyxy ydx

= + .

4 (a) If 2y ax= b+ write down dydx

.

(b) Determine ( )345 25x x

dx+

d

.

5 (a) Determine the gradient of the curve 24 3y = + x at the point where x = 2 . (b) Differentiate 2 336 3 5 2 6x x= + + + +y x . x

)

6 Differentiate each of the following with respect to x: (a) ( ) (b) (2 23 2 4 3x x x x+ − + − ( )( )2 22 3 4x x x+ − + x

)

(c) ( ) (d) (3 2 2 7 4x x x x x+ + + + ( )( )2 29 4 7x x x− + + (e) ( )(3 29 2 2 4 7 )3x x x− − − x (f) ( )( )2 22x x x+ + (g) ( ) (h) ( ) (21 2 3x x+ + )2 )

)4

(2 23 2 4 1x x+ −

(i) ( ) (j) (22 31x x+ + ( ) ( )2 22 2 5x x x− +

(k) ( ) (l) (1 21 x x−+ )2+ 1x x+ (m) ( ) (n) ( ) (11x x−+ − )3 )(1 13 2 3x x− −− + (o) 3) 3( 4x x+ + x (p) 24) 1x x+ −(3 7 Differentiate each of the following with respect to x:

(a) 1

xx +

(b) 1

xx −

(c) 4 (d) 79 3

xx++ 2

11x

x +−

(e) 2

2 1x

x − (f) 22

x6 5x−−

(g) 31

xx−

(h) 2 3 13 1

x xx− −−

(i) 21x

x+ (j) 2

5 xx−

(k) 2

24

3 53x x

x−

− (l)

2

22 31 2x x

x x+ −+ +

(m) 21)

1x2(

x x+

+ − (n) 1

1xx

+−

(o) 11

xx

−+

(p) 21xx−

(q) 212x

x+ (r)

2

11 x

x−+

8 Differentiate each of the following with respect to x:

(a) 2

11 x

x+−

(b) 2

( 2)( 3x

x x+ + )

(c) 3

3 1x

x + (d)

3

311 x

x+−

(e) 2

211

x xx x

− −+ +

(f) 23 1x −

(g) 21 51

x− (h) 2

1xx++

(i) 510x

x + (j) 2 1

3 4xx−+

(k) 1

xx +

(l) 1xx−

(m) 11

xx−+

(n) 22 x x

x+ +

(o) 4

212x

x− (p)

2

23 1x xx x− −− +

2 3 4

(q) 1/ 2

3/ 22x

x+ (r)

3/ 2

1/ 23 x

x−

(s) x ax a+− (t) x a

x a+−

(u) ( 12x

)x x −−

(v) 3

2 2a x−x

(w) 2( 2)

5 4x x+

− +x x (x)

3

311x +

x −

(y) 51

xx −

(z) 31

xx +

9 Find dydx

for:

(a) 2 5

xx +

y = (b) 23 2

xy = x +

(c) xax b+

y = (d) 23 2

xx−

(e) 2

xax b+

y = (f) 3 22 54 3

xx

y x += + −

+

Exercise 3 - Differentiation of Exponential Functions 1 Differentiate the following with respect to x: (a) 6xe (b) 8xe (c) xe− (d) 2xe− (e) 3xe (f) + 4xe + (g) 2xe− + (h) 3xe− + (i) 4 xe (j) − 5 xe − (k)

2xe (l) 3xe

(m)

2xe (n) − 3xe− (o) 8 5xe − (p) 9 4xe − (q) 7 3xe (r) + 8 9xe + (s)

2 2x xe + (t) 23 5x xe +

(u)

2 3 1x x− +e (v) 2 32 xxe − + (w) xe (x) 2 xe (y) 3xe (z) + 4xe + 2 Differentiate the following with respect to x: (a) 4 xe (b) − 5 xe − (c) (d)

2( 2)xe + 2( 3)xe −

(e) e (f) (g)

2( 1)x− 2( 2)xe − xx e (h) 32 xx e

(i) 3 2xx e− (j) xx e− (k) 2xex

(l) 3xex

(m) xe

x

−

(n) 2xex

−

(o) 2 xx e (p) 2xx e

(q) 3xx e (r) 3 xx e (s) xx e−

(t) (u) 3 2xx e (v) 35 xx e

3 Find the derivative of: (a) ( 2)x xe e−+ (b) 2( )x xe e−− (c) 2( )xx e− (d) ( 2)xx e−+

(e) 2 2( )2x xe e−− (f) 2 2( )2x xe e−+ (g) ( 1) xx e+ (h) ( 1) xx e−−

(i) 3( 3) xx e−− (j) ( 4) xx e+ (k) 1

xex +

(l) 1

xex

−

−

(m) 1x −e (n) 2 1xe +

4 Determine dy given that: dx

(a) 2 3( )x xe e−= − xy e (b) 2x x

xe e

e

−+y =

(c) 3 2x x

xe

e

−

−−e (d) 2 2( 1)xy e= +

(e) 2 2( 2)x xe−= +y e (f) 2

3 21 ( )x

xe− = + +

y x

(g) 2( x= + 3)y e (h) 3 4(2 5)xy e = + 5 Differentiate each of the following with respect to x : (a)

2(1/ 2) xe (b) − 2ex

Exercise 4 - Differentiation of Logarithmic Functions 1 Find the derivative of: (a) log 5e x (b) loge 3x (c) log 6e x (d) loge 7x (e) 2 log 3e x (f) 3 loge 4x (g) l (h) l og ( 1)e x − og (2e x −1) (i) lo (j) lo2g ( 3)e x + 2g (4 )e x− (k) 2 loge x (l) 3 loge x

(m) 1gelox

(n) 2gelox

(o) (l 4og )e x (p) (l 3og )e x (q) 2g (2 3)e x +lo (r) 2g (3 1)e x −lo (s) 2g ( 4)e x x+ −lo (t) 2g ( 1 )elo x x− −

(u) lo (v) 2 2g ( 2)( 3)e x x+ +2 1g

3elo x xx+ −−

2 Differentiate the following with respect to x: (a) lo2 gex x (b) lo 2gex x (c) lo (d) g (3 4)e x − log (2 5 )e x−

(e) 3log3e

xx

−+

3 Find the derivative of each of the following:

(a) 23 1x

x−+

y =

(b) y 3log ( 1)e x= + (c)

23 54 2x x− +=y e

(d) 2( ) ( 1)( 5 3)f x x x x= − + + (e) 2( ) ( 3)xg x e x= +

(f) 5

3(2 3)( )

5xg x

x+

=−

(g) [ ]4( ) log (5 1)e x= −f x

(h) loge xyx

=

(i) 3/ 2

2log( ) e xg x

x=

Exercises A - Answers

Exercise 1 1 (a) 12 (b) 1 (c) 5 (d) 2

1 7

(e) 2 3x (f) 53 (g) 10 (h) 2a

(i) -5 (j) 13 (k) 27 (l) 3 2a

(m) 5 22 1x x− + (n) 12 (o) 0 (p) 1

3 2 (a) 2 (b) 2 12x − x + 3 (a) 6x (b) 6x (c) 5 (d) -3 4 (a) 4 5 2x h− + (b) 4 5, 3x − 5 (a) 2 8x − (b) when 4x = gradient is 0

6 (a) 8 (b) 1x + 2 2

1x−

7 (a) 4 3x − (b) 6x − 8 (a) 4 7x− (b) 4 3+ (2 )x + 10 (a) h (b) 0 11 (a) 8 (b) 8 h+

12 (a) 32

x− (b)

3

12 x− (c)

3/ 2

3x−2

(d) 3/ 4

4

−x (e) 54

x− (f)

4/3

3x−−

(g) 5/32

3x−− (h) 200 199x (i) 2.4

1.4x−

(j) 1.90.9

x− (k)

8/353x−− (l)

p qqp x

q

−

13 (a) 4 3x − (b) 82x−+ (c) 2 b+ ax

(d) 6 1 (e) 2 2x x− + 6 2 1x x− + (f) 3 2 + 2ax bx c+ +

(g) 3 24 3 2x x − c 3− − (h) 8 13 22 4x x x− + − 14 (a) 2 3x − (b) 2

15 (a) 6 62 5x x− + (b) 12 18x− (c) 211x

−

(d) 32x2x

− (e) 6 4(3 )x − (f) 241x

−

(g) 283

3x x

− (h) 3/ 22

3

22

x3 x x−− −

16 3/ 2 1/ 2 1/ 29 1 1

2 25 ,x x x−− + 431 18 (2 , 2) 19 (a) -29 (b) 6 20 (a) 0 (b) (2 , -1)

22 3 (2 1)8

xx−

Exercise 2

1 (a) 52 9 5x

− (b) − 32 (3 4)x −

3− (c)

32 (6 5 )5

x−

(d) 36

(2 3 )x− (e)

32 (7 8)x +7

−

(f) 3 ( (g) 2 (

(h)

3 2 2 21) (3 2 1)x x x x x− + − − + 3 2 21)(3 2 1)x x x x x+ + − + +

3

52 (5 7)x −

− (i) 7 ( 2 63 15 ) (6 15)x x x− −

(j) 2

22

1 1x x x3 2 1x x

+ + + −

(k) 2 8x

x + (l)

28x

x−

− (m)

28 78xx −

(n) 2

3

127 8

xx

− (o) − 22 3

xx x2 3

7+

+ − (p)

22 4 3 9x

x x8 3++ −

(q) 2

2 12 1

xx x

+

+ − 0 (r)

2

113 12 11

x6 1x x

−

+ − (s)

2 3( 10)x +

x−

(t) 2 3(9 )

xx−

(u) 2(3 2 1)

xx x 3

3 1+− (v)

+ + 2 3(7 6 5)x

x x7 3−

− − +

2 (a) 23

23 ( 1)

xx + 2

(b) 2 455 ( )

xx x

1 2−−

(c) 34 (7 )x− −

(d) 2 760

(1 5 )xx

− (e) +

6

21 1x 135x x

− +

(f) 2 2(1 )6xx

− +

(g) 23

(7 3 )x− (h)

412

5 ( )xx

+ (i) 2 2(2 )4x

x−

(j) 60 3 4(1 )4x x− (k) 21

4xx+

(l) 3 3 (1 3 )

4x

x+

(m) 3 7 (1 7 )

4x

x+

(n) 2 4(7 3 )18x

x− (o) 1

2 13 x−

−

(p) 424

(7 8 )x− (q)

+45 2

4 1 3 (6x x x 1)− + − (r) 23 (1 )x x− −

(s) 18 (t) 8(2 3)x −4

4 3105 11x x

+

− (u)

2

3

3(1 )2 1 3

xx x− −

− +

(v) 3 ( 2 29 5 2 ) (5 4 )x x x+ + +

4 (a) 2

axax b+

(b) 4 5 234 (5 10 ) ( 5 )x x x x 1/ 4( ) −+ +

5 (a) 3

2 (b) 12 2 33 2( 1) (2 6 )x x x x 2/3−+ + + + 6 (a) 4 2 (b) 8 13 21 14 1x x x+ + − 7 43 25 2x x x+ − −

(c) 5 3 (d) 4 3 22 36 22x x x x+ + + + 4 33 24 21 10 6x x x− − + +

(e) 84 (f) 4 35 4 3 240 252 120 18x x x x+ − − + 3 2 4x x x 2+ + +

(g) 2 (h) ( )( )(1 2 3 4 5x x x+ + + ) ( )( )( )2 4 1 24 5x x x+ − +2 3

(i) x x (j) ( )(2 31 7 3 16x x+ + + ) ( )( )( )22 5 6 6x x x x x− + + −2 1 5

(k) 2

22

(1 )xx

+ −+

2x (l) 3 22 1

xx

++

(m) 2( 1)x +4

(n) 26 7

(3 2) ( 3)x

x x+

− + 2− (o) 3 2

2

2 9 12 3

x x xx x

5 1 2+ + +

+ (p)

2

2 1x x

x+ −

−

6 4 3

7 (a) 21

( 1)x + (b) 2( 1)x −

1− (c) 2(9 3)x +

51−

(d) 21

( 1)x −− (e) 2( 1)2

2xx −

− (f) 2

2 25 6)

(2 )x x

x− +−

2(3

(g) 23

(1 )x− (h)

2

2(3 1)x x

x3 2 6− +

− (i) 3

42 1

x3x x

+−

+

(j) 30

2 5x3 2

x x−−

(k) 2

2 210 9 6)(3 5 )

x xx

2 ( − +−

(l) 2

2 25

( 3 2)1 6x xx x− −+ +

(m) 2

22 2)

( 1x x

x x+ ++ − 2)

2 (− (n)

3

11 (1 )x x+ −

(o) 3

1(1 ) (1 )x x− +

−

(p) 2

2 2(1 )1 x

x+−

(q) 2

2 21

(1 )2 ( )x

x−

+ (r) -1

8 (a) 2

2(1 )1 2x x

x+ −−

(b) 2(5 12)

( 5 6)x x 2x x +

− (c) + +

2

3 2( 1)x

x +3

(d) 2

3 2(1 )x6x−

(e) 2)

( 1x x

x x 2)2 ( 2++ +

(f) 2(3 1)x −6

−

(g) 2 210

(1 5 )xx−

(h) 2( 1)x +1

− (i) 2( 10)x +50

(j) 211

(3 4)x + (k) 22 ( 1)x x

1 x−+

(l) 3

12x

x+

(m) ( )2

1

1x x + (n) 2 1

x2

− + (o) 3 2

2 22 )

(1 )4 (x x

x−

−

(p) 2

24 1)

(3 1)x xx x+ −− + 2

7( (q) 2 5

1 3x x

− − (r) 3

3 12 x

− −

(s) 22

( )a

x a+ (t) 2( )

a2x a−

− (u) 2

24 2

( 2)x

x− +−

x

(v) 2 2 2

2 2 2(3 )

( )x a x

a x−

− (w)

2

27

( 5 4)28 8x xx x+ −− +

(x) 2

3 3( 1)6x

x +

(y) 25

( 1)x −− (z) 2( 1)x +

3

9 (a) ( 5 (b) (33/ 2)(2 5)x x −+ + 3/ 24)(3 2)x x −+ + (c) 3/ 2

2 )( )ax b ax b1 ( 2 −+ +

(d) 3 ( (e) b a2 3/ 23 2 )x −− 2 3( )x b / 2−+ (f) 27

2(4 3) 3 2x

x x122 +

++ +

Exercise 3 1 (a) 6 6xe (b) 8 8xe (c) xe−−

(d) 22 xe−− (e) 3xe + (f) 4xe +

(g) 2xe− +− (h) 3xe− +− (i) 4 xe −−

(j) 5 xe −− (k) 22xxe (l) 3

32 xx e

(m) 2

2 xxe−− (n) 323 xx e−− (o) 8 55 xe −−

(p) 9 44 xe −− (q) 3 7 3xe + (r) 9 8 9xe +

(s) 2 (2 21) x xx e ++ (t) (6

2 535) x xx e ++ (u) (22 3 13) x xx e − +−

(v) 2 (2 2 31) x xx e − +− (w) 1 (x)

2xe

x2

x1 xe

(y) 312 3

xex

+

+ (z) 4

2 4xe

x1 +

+

2 (a) 412 4

xex

−

−− (b) 51

2 5xe

x−− (c) 2(

−

2( 2)2) . xx e ++

(d) 2 ( (e) 2 ( (f) 2 ( 2( 3)3) . xx e −−

2( 1)1) . xx e −−2( 2)2) . xx e −−

(g) ( 1) xx e+ (h) 2(3 31) xx e+ (i) 3( 2 21 ) xx e−−

(j) (1 ) xx e−− (k) 2

21)(2 xx e (l)

x− 3

21)(3 xx e

x−

(m) 2(1 ) xx e

x

−− + (n) 2

2(1 2 ) xx e

x

−− + (o) ( 2) xx x e+

(p) (1 22 ) xx e+ (q) (1 3 3) xx e (r) ( 3+ 2 ) xx x e+

(s) (1 2 )2

xx ex

−− (t) 21)

2(4 xx e

x+ (u) 3( 22 1) xx e+

(v) 5 ( 33 1) xx e+ 3 (a) 2 ( 2 2 )x xe e−− (b) 2 ( 2 2 )x xe e−− (c) 2 ( )(1 )x xx e e−−

(d) 2 ( )(1 )x xx e e−+ − − (e) 4 ( 4 4 )x xe e−− (f) 4 ( 4 4 )x xe e−−

(g) ( 2) xx e+ (h) (2 ) xx e−− (i) (1 3 30 ) xx e−−

(j) ( 5) xx e+ (k) 2( 1)

xxex +

(l) 2)

( 1)( 2 xx e

x−−

(m) 2 1

x

xe −

e (n) 2

2 1

x

xee

+

4 (a) 3 3 2 2x xe e−+ (b) 22x xee −− (c) 4 4xe

(d) 4 4 4 2x xe e+ (e) ( )4 44 x xe e−− (f) 63

22 6 xx ex

−− −

(g) 2

2 3

x

xe +

e (h) 24 3 3(2 5)3x xe e +

5 (a)

21/ 2 xxe−− (b) e x22 e 1−

Exercise 4

1 (a) 1x

(b) 1x

(c) 1x

(d) 1x

(e) 2x

(f) 3x

(g) 11x −

(h) 22 1x −

(i) 2 32x

x +

(j) 22

4xx

−−

(k) 1x

(l) 32x

(m) 1x

− (n) 12x

− (o) 3log )ex4 . ( x

(p) 2og )ex3 (l x (q)

2 2

2

(2 3) log (2 3)e

x

x x+ +

(r) 2

3

(3 1) log (3 1)e

x

x x− −2 (s)

2 4x1−

(t) 2

2 2

1

1 1

x x

x x x

+ − − − −

(u) 2

2 22 )

( 2)( 3x x

x x+2 ( 5

)+ +

(v) 2

26 2

( 1)(x x

x x x− −

+ − −3)

2 (a) (1 2 log )ex x+ (b) 2 l(1 og )e x+

(c) 32 (3 4) log (3 4)ex x− −

(d) 52 (5 2) log (2 5 )ex x− −

(e) 3( 3)( 3x x− + )

3 (a) 3

3 82 (3 1)

xx+

+

(b) 31x +

(c) 4 (23 56 5) 2x xx e − +− (d) 3 82x x 2+ −

(e) e x (f) ( 2 2 3x x+ + )3 2

3 24 9 50)(2 3)

( 5)x x x

x− − +

−

4(

(g) 320 [log (5 1)]

5 1e xx

−−

(h) 2log

2e1 2 x

x−

(i) 3/ 2

33 4 log

2e x

x−

Calculus - A Basic Rules

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Transcript of Calculus - A Basic Rules