Differentiation of Exponential Functionstlment.nait.ca/.../C33-Differentiation_of_Exponential_Functions.pdf · Module C33 − Differentiation of Exponential Functions EXAMPLE 1 Find

LAST REVISED March, 2009

Copyright This publication © The Northern Alberta Institute of Technology 2007. All Rights Reserved.

Calculus Module C33

Differentiation of Exponential Functions

Module C33 − Differentiation of Exponential Functions

Introduction to Differentiation of Exponential Functions Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material?

Learning Outcome When you complete this module you will be able to…

Learning Objectives 1. Find the derivative of the exponential function uy b= where ( )u f x= . 2. Find the derivative of where uy e= ( )u f x= . 3. Use the chain rule when finding the derivative of exponential functions. 4. Find the derivative of product functions involving exponentials. 5. Find the derivative of quotient functions involving exponentials. 6. Find the derivative of exponential functions containing trigonometric relations. 7. Find successive derivative of exponential functions. 8. Find the slope of an exponential curve at a given point. 9. Find maximum, minimum and/or inflection points for exponential functions.

Connection Activity

1

http://www.microsoft.com/windows/ie/default.asp

http://www.macromedia.com/shockwave/download/download.cgi?P1_Prod_Version=ShockwaveFlash


OBJECTIVE ONE When you complete this objective you will be able to… Find the derivative of the exponential function uy b= where ( )u f x= .

Exploration Activity Formula Derivation Begin with

( )ln xe x=

Differentiate both sides of this equation

( )1 1xx e

e′ =

Multiply by xe

( )x xe e′ =

Now apply the chain rule to obtain

( )u ud de edx dx

=u

Now for other bases we use the fact that

ln aa e= Therefore

( ) ( )( ) ( ) ( )ln ln ln lnln ln lnx a x a x a x aa e x e e x a a e a′′ ′ ′= = = = × = xa×

Now apply the chain rule to obtain

( ) lnu ud dua adx dx

= i a (Formula 1)

2 2


EXAMPLE 1 Find for y′ 4 110 xy += SOLUTION:

( )4 1 4 1ln10 10 4 1 4ln10 10x xy x+ +′′ = × × + = ×

EXAMPLE 2 If 3 42 xy += find dy dx/ and evaluate it at both 1x = − and 0x = . SOLUTION: Using formula 1 from the previous page:

( )3 4 3 4

4

ln 2 2 3 4 3ln 2 2

( 1) 3ln 2 2 4.1589 (0) 3ln 2 2 33.2711

x xy x

y y

+ +′′ = × × + = ×

′ ′− = × = = × =

3


Experiential Activity One Find the derivative of the following functions: 1. (3 2)5 xy +=

2. ( 3)3 xy −=

3. 22 3 14 x xy − + +=

4. 2( )5 x xy −=

5. 5(4 3 )7 xy −=

6. 610 xy = For the following find dy dx/ ; also find the value of dy dx/ at the given value for x. 1.

2

3 1xy x= ; =

2. 2( 2 1)10 1x xy x+ += ; = −

7.

3

1.

3. 2

4 1xy x−= ; = −

4. 3( 1)6 1xy x−= ; =

5. 242 0xy x= ; = .

6. 3(1 2 )5 0x xy x− −= ; =

Experiential Activity One Answers 1. 3 23(5 )(ln 5)x+

2. 3(3 )(ln 3)x−

3. 22 3 1( 4 3)(4 )(ln 4)x xx − + +− +

4. 2

(2 1)(5 )(ln 5)x xx −−

5. 54 4 3( 15 )(7 )(ln 7)xx −−

6. 6(6)(10 )(ln 10)x

7. 6 5917.8. 0 0000.9. 0 6931.10. 17226 8015.11. 2 1350.12. 11 8208− .

4 4


OBJECTIVE TWO When you complete this objective you will be able to… Find the derivative of where . uy e= ( )u f x=

Exploration Activity If the function has the natural base, e, then formula 1 from objective 1 can be used. We simply replace the base b with e and the formula becomes;

( ) ( )ln

u

u

y edy due edx dx

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

but notice, if you take the natural log of base the result is 1. eTherefore the formula for finding the derivative of an exponential function can be shortened to:

if uy e=

then ( )udy duedx dx

⎛ ⎞= ⎜ ⎟⎝ ⎠

(Formula 2)

The two steps are:

1. The first part of the derivative is the same as the function. 2. Then multiply by the derivative of the exponent.

EXAMPLE 1 If 4 3xy e −= , find dy dx/ . SOLUTION:

( )( )4 3

4 3

(function) (derivative of the exponent)

4

4

x

x

dydx

e

e

−

−

= ⋅

=

=

5


EXAMPLE 2 If

2 4 1x xy e − += , find dy dx/ and evaluate it at 0x = and at 2x = . SOLUTION:

( )( )2 4 1

1

3

2 4

at 0, ( )( 4) 10 8731

at 2, ( )(0) 0

x xdy e xdx

dyx edxdyx edx

− +

−

= −

= = − = − .

= = =

6 6


Experiential Activity Two Find dy dx/ for each of the following: 1. (7 11)xy e −=

2. 3 1xy e −=

3. 23xy e=

4. xy e=

5. 23 5 2x xy e + −=

6. 5xy e=

7. If 25 6 3x xy e − −= find dy dx/ and evaluate the derivative at 0 90x = .

8. If find 26 2 23x xy e + −= dy dx/ and evaluate the derivative at 2 0x = .

9. If 5 6xy e −= find dy dx/ and evaluate the derivative at 1 50x = .

10. If find 28 1xy e −= dy dx/ and evaluate the derivative at 0 70x = .

11. Find dy dx/ and evaluate it at for: 0 7x = .28 4 9x xy e − −=

12. Find dy dx/ and evaluate it at for: 1 9x = .25 5 26x xy e + −=

Experiential Activity Two Answers 1. 7 117( )xe −

2. 3 13( )xe −

3. 236 ( )xx e

4. 1 21 ( )2

xx e− /

5. 23 5 2( 4 5)( )x xx e + −− +

6. 55( )xe 7. 0.038720 8. 3858 7421.9. 22 4084.10. 207 6624.11. 0.002723 12. 113 0753.

7


OBJECTIVE THREE When you complete this objective you will be able to… Use the chain rule when finding the derivative of exponential functions.

EXAMPLE 1 If , find

3(2 1)xy e += y′ SOLUTION: Use formula 2 from objective 2 and the chain rule.

3

3

(2 1) 2

2 (2 1)

(3)(2 1) (2)

6(2 1)

x

x

y e x

x e

+

+

′ = +

= +

EXAMPLE 2 If , find and evaluate it at

2 3(1 )10 xy −= y′ 1x = . SOLUTION:

( ) ( ) ( )

( ) ( )

32

32

21 2

2 12

ln10 10 3 1 0 2

6 1 ln10 10

x

x

y x

x x

−

−

′ = × × − × −

= − − × ×

x

Evaluating at x = 1

0y′ =

8 8


Experiential Activity Three 1. Find dy dx/ for:

2( 48)xy e +=

2. Find the derivative of ( )225 30xy e −=


4. Given find 2 3(9 65)xy e −= y′


6. Given find 3(6 16)xy e −= dy dx/ and evaluate it at x = 2.9.

7. Given find 2 3(5 30)xy e −= dy dx/ and evaluate it at x = 2.5.

8. Given find 2(2 4)xy e −= dy dx/ and evaluate it at x = 2.9.

9. Given find 2( 3 2)x xy e − +=

2

dy dx/ and evaluate it at x = 1.7.

10. Given find 2 4(3 )10 xy = dy dx/ and evaluate it at x = 0.6.

Experiential Activity Three Answers 1. ( ) ( )2482 48 xx e ++

2. ( ) ( )225 30220 5 30 xx x e −−

3. ( ) ( )222 1328 2 13 xx x e −−

4. ( ) ( )229 65254 9 65 xx x e −−

5. ( ) ( )29 1018 9 10 xx e −−

6. ( ) ( )32 6 16

2.918 6 16 548.5707x

xx e −

=− =

7. ( ) ( )322 5 302

2.5

30 5 30 826.2523x

x

x x e −

=

− =

8. ( ) ( )22 4

2.94 2 4 183.8428x

xx e −

=− =

9. ( )( ) ( )22 3 22

1.7

2 2 3 3 2 0.175575x x

x

x x x e − +

=

− − + = −

10. ( ) ( )423 32

0.6

ln10 4 3 6 957.9402x

x

x x e=

× × × =

9


OBJECTIVE FOUR When you complete this objective you will be able to… Find the derivative of product functions involving exponentials.

Exploration Activity For some problems it will be necessary to apply the product rule of derivatives. The product rule states: If where u and v are functions of x then: y u v= ⋅

dy dv duu vdx dx dx

= ⋅ + ⋅

A review of this concept may be found in module C12 – Derivatives by Formula

EXAMPLE 1 If 2 cos xy x e= , find dy dx/ . SOLUTION: Use formula 2 from objective 2 and apply the product rule.

2 cos

2 cos cos

cos

and

( )( )( sin ) ( )(2 )

( sin 2)

x

x x

x

u x v edy x e x edx

xe x x

= =

= − +

= − +

x

EXAMPLE 2 If 3 32 xy x e−= , find dy dx/ and evaluate it at x 1 5= . . SOLUTION:

3 3 3 2

2 3

(2 )( )( 3) ( )(6 )

6 ( 1)

x x

x

dy x e edx

x e x

− −

−

= − +

= − −

x

and at x = 1.5 we get = −0.074986

10 10


Experiential Activity Four

11

8)1. Given find 3 (4 0(7 8) xy x e + .= − dy dx/ and evaluate it at 0 4x = . .

2. Find the derivative of the following function and evaluate it at 0 6x = . : 2 (2 0 4)7 xy x e − .=

3. Find the derivative of the following function and evaluate it at 0 1x = . :

2 (5 0(6 7) xy x e + .= + 3)

4. Find the derivative of the following function and evaluate it at 0 6x = . :

2 (2 0 8)5 xy x e + .= 5. Find the derivative of the following function and evaluate it at 0 5x = . :

2 (4 0(3 7) xy x e − .= + 1)

4) 6. Given find dy3 (5 0(8 6) xy x e − .= − dx/ and evaluate it at 0 3x = . .

Experiential Activity Four Answers 1. ( )( ) ( )4 0.82 3

0.421 4 7 8 295.9502x

xy x x e +

=′ = + − = −

2. ( ) ( )2 0.42

0.614 29.9113x

xy x x e −

=′ = + =

3. ( )( ) ( )5 0.32

0.112 5 6 7 81.2322x

xy x x e +

=′ = + + =

4. ( ) ( )2 0.82

0.610 70.9349x

xy x x e +

=′ = + =

5. ( )( ) ( )4 0.1

0.56 4 3 7 227.3204x

xy x x e −

=′ = + + =

6. ( )( ) ( )5 0.42 3

0.324 5 8 6 80.3915x

xy x x e −

=′ = + − = −


OBJECTIVE FIVE When you complete this objective you will be able to… Find the derivative of quotient functions involving exponentials.

Exploration Activity Sometimes it is necessary to apply the quotient rule of derivatives. The quotient rule states:

If = y uv

where u and v are functions of x then

12 12

2

du dvv udy dx dxdx v

−=

A review of this concept may be found in module C12 – Derivatives by Formula

EXAMPLE 1

If 4

4

xeyx

= , find dy dx/ .

SOLUTION: Use formula 2 from objective 2 and apply the quotient rule.

( )

4 4

4 4 4 3

8

4

5

and( )( )(4) ( )(4 )

4 ( 1)

2

x

x x

x

x

u e v xdy x e e xdx x

e xx

x xe

= =

−=

−=

−=


EXAMPLE 2

If 2

x

xye

= , find dydx

and evaluate it at 0 5x = . .

SOLUTION:

2

2

2

2

2

and( )(2 ) ( )( )

( )(2 )

(2 )

x

x x

x

x

x

x

x

y x v edy e x x edx e

xe xe

dy xe xdx e

= =

−=

−=

−=

and at x = 0.5,

0 50 5 (2 0 5) 0 45492 7183

dy edx

.. − .= =

..

It is easier to avoid the quotient rule as follows:

( )2

2 22 2

x

x x x

y x e

y xe x e x x e

−

− −

= ⇒

′ = − = − −

13


Experiential Activity Five

1. Find dy dx/ and evaluate dy dx/ at 0 7x = . for: (5 8)

2

xeyx

−

=


6x

xye −=


5

xeyx

+

=


5 2

xeyx

+

=+

5. Find the derivative of 6

433

xey ex

= − then evaluate it at 0 1x = . .

6. Find the derivative of 5

424

xey ex

= + then evaluate it at 0 5x = . .

Experiential Activity Five Answers

1. 5 8 5 8 5 8

2 20.7

5 (2 ) 2 (5 1) 0.028344 2

x x x

x

e x e x eyx x

− − −

=

− −′ = = =

2. 1 2

1 2

0.3

6

(6 12 ) 3.5804

x

x

x

y xe

y x e

−

−

=

=

′ = − =

3. 4 7 4 7 4 7

2 20.5

20 5 (4 1) 6482.467125 5

x x x

x

xe e x eyx x

+ + +

=

− −′ = = =

4. 8 2 8 2 8 2

2 20.4

8(5 2) 5 (40 11) 305.8969(5 2) (5 2)

x x x

x

x e e x eyx x

+ + +

=

+ − +′ = = =+ +

5. ( )

4 3 6

4 3 6

0.1

13

1 3 6 14576.95043

x

x

x

y e x e

y x x e

−

− −

=

= −

′ = − + =

6. ( )

4 2 5

3 2 5

0.5

14

1 2 5 12.18254

x

x

x

y e x e

y x x e

−

− −

=

= +

′ = − + =

14 14


OBJECTIVE SIX When you complete this objective you will be able to… Find the derivative of exponential functions containing trigonometric relations.

Exploration Activity You will need to refer to the rules for finding the derivatives of the trigonometric functions.

EXAMPLE 1 If tan xy e= , find dy dx/ . SOLUTION: Use formula 2 from objective 2 and remember the derivative of tan x is 2sec x.

tan 2( )(sec )xdy e xdx

=

EXAMPLE 2 If sin 310 xy = , find dy dx/ and evaluate at 6x π= / . SOLUTION:

Use formula 1 from objective 1, and remember you must find the derivatives of sin 3x

sin 3(10 )(3cos 3 )(ln 10)

, 06

xdy xdx

dyxdx

π

=

= =

15


Experiential Activity Six 1. If sin 7 xy e= find dy dx/ and evaluate the derivative at 0 7x = . .

2. If sin 7x xy e= find dy dx/ and evaluate the derivative at 0 1x = . .

3. If find 9 arctan9xy e x= dy dx/ and evaluate the derivative at . 0 5x = .

4. If find 3 cos (8 )xy e x= dy dx/ and evaluate the derivative at . 0 4x = .

5. If 9

8sinx

xye

= find dy dx/ and evaluate the derivative at 0 22x = . .

Experiential Activity Six Answers 1. sin 7

0.77cos7 0.4888x

xy xe

=′ = =

2. ( ) sin 7

0.1sin 7 7 cos7 1.2581x x

xy x x x e

=′ = + =

3. 92

0.5

19 arctan 9 1133.55661 81

x

x

y x ex =

⎛ ⎞′ = + =⎜ ⎟+⎝ ⎠

4. ( ) 3

0.43cos8 8sin8 8.3929x

xy x x e

=′ = − = −

5. ( )

9

90.22

8sin1.0915

8 cos 9sin

x

xx

y xey x x e

−

−

=

== −

′ = −

16 16


OBJECTIVE SEVEN When you complete this objective you will be able to… Find successive derivatives of exponential functions.

Exploration Activity If can be useful in certain applications to find the second and third derivatives of exponentials. It is sometimes easier to find the second derivative to test maximum and minimum points. Also the test of an inflection point can come from the third derivative of a function.

EXAMPLE 1 If 10xy e−= , find the second derivative. SOLUTION:

10

10

210

2

10

( )( 10)

10

and 10 ( 10)

100

x

x

x

x

dy edx

ed y edx

e

−

−

−

−

= −

= −

= − ⋅ −

=

EXAMPLE 2 If

2xy e= , find the second derivative and evaluate it at 1 5x = . . SOLUTION:

2

2

(2 )

2

x

x

dy e xdx

xe

=

=

Now to find the second derivative we must apply the product rule:

2 2

2 2

2

2

2

2

2

(2 )( )(2 ) 2

4 2

2 (2 1)

x x

x x

x

d y x e x edx

x e e

e x

= +

= +

= +

17


Experiential Activity Seven 1. Find the second derivative of the following function and evaluate it at t = 0.7:

3( 8)ty e= −2. Find the second derivative of the following function and evaluate it at x = 0.67:

3 sin 7xy e x=3. Find the second derivative of the following function and evaluate it at x = 0.63:

arctan xy e= 4. Find the second derivative of the following function and evaluate it at x = 0.51:

4 lnxy e x= 5. Find the second derivative of the following function and evaluate it at x = 0.5:

45 xy x e= 6. Find the second derivative of the following function and evaluate it at x = 0.67:

3 cos 3xy e x=

Experiential Activity Seven Answers

1. ( )( ) ( )

2

2 22

0.7

3 8

6 8 3 8 70.8369

t t

t t t t

t

y e e

y e e e e=

′ = −

′′ = − + − =

2. 3

3

0.67

(3sin 7 7cos7 )

( 40sin 7 42cos7 ) 291.4404

x

x

x

y e x x

y e x x=

′ = +

′′ = − + =

3. ( )( ) ( )

2

2 2 3

2 22 2

0.63

1

1 20.2315

1 1

x

x

x x x x x x

x x

x

eye

e e e e e eye e

=

′ =+

+ − −′′ = = =+ +

−

4. 4 1 4

4 1 4 2 4

0.51

4 ln

16 ln 8 8.2142

x x

x x x

x

y e x x e

y e x x e x e

−

− −

=

′ = +

′′ = + − =

5. 4

4

0.5

(5 20 )

(40 80 ) 591.1245

x

x

x

y x e

y x e=

′ = +

′′ = + =

6. 3

3

0.67

3 (cos3 sin 3 )

18 sin 3 121.5896

x

x

x

y e x x

y e x=

′ = −

′′ = − = −

18 18


OBJECTIVE EIGHT When you complete this objective you will be able to… Find the slope of an exponential curve at a given point.

Exploration Activity As stated earlier, the result of finding the derivative of a function represents the formula for finding the slope of a curve at a given point. Therefore, find the derivatives of any function, then substitute a particular x -value and you will find the slope of the function at that point. Module 13 will provide you with a review of this concept.

EXAMPLE 1 If

2xy e= , find the slope of the curve at 1x = . SOLUTION:

2

2

2

2

x

x

dy e xdx

xe

= ⋅

=

at x = 1,

1(2)(1)( ) 5 4366dy edx

= = .

at x = 1, substituting into the original equation, y = 2.7183. Therefore at the point (1, 2.7183), the slope is: 5.4366.

19


EXAMPLE 2

If 3

3

xeyx

= , find the slope if x = 2.

Also, does the curve have a horizontal tangent, if so, find the point. What is the slope at that point?

SOLUTION:

3 3 3 2

6

2 3

6

3

4

3 3

3 ( 1)

3 ( 1)

x x

x

x

dy x e e xdx x

x e xx

e xx

⋅ ⋅ − ⋅=

−=

−=

at x = 2,

63 (1) 75 6429

16dy edx

= = .

Therefore at the point (2,50.43) the slope is 75.6429 Now, for a horizontal tangent the slope must be zero. Therefore set the derivative equal to zero and solve for x.

3

4

3 ( 1) 0xe xx−

=

multiply both sides by 4x 3(3 )( 1) 0xe x − = set both factors to zero 33 0 or 1xe x= − = 0 3 0 or 1xe x= = The factor 3xe is always positive, therefore it can never be equal to zero. So

1x = is the only acceptable value. At represents the point where the curve has a horizontal tangent and where the slope is zero.

1 20x y= , = .09

20 20


Experiential Activity Eight 1. If find the slope at

2cos( 6)xy e −= 3 20x = .

2. If ( 2)

3

xeyx

−

= find the slope at 6 00x = .

3. If sin 7

9

xeyx


4. If (12 9)

23

xeyx

−


5. If 62

1xxy e

e= − find the slope at 0 35x = .

Experiential Activity Eight Answers

1. ( ) ( )2cos 62

3.22 sin 6 3.6157x

xy x x e −

=

′ = − − =

2. ( )3 2

4 3 2

63 0.1264

x

x

x

y x e

y x x e

− −

− − −

=

=

′ = − + =

3. ( )

1 sin 7

2 1 sin 7

1

191 (7cos7 ) 0.91689

x

x

x

y x e

y x x x e

−

− −

=

=

′ = − + =

4. ( )

2 12 9

3 2 12 9

1

131 2 12 66.98183

x

x

x

y x e

y x x e

− −

− − −

=

=

′ = − + =

5. 6 2

6 2

0.356 2 49.9902

x x

x x

x

y e e

y e e

−

−

=

= −

′ = + =

21


OBJECTIVE NINE When you complete this objective you will be able to… Find maximum, minimum and/or inflection points for exponential functions.

Exploration Activity Maxima or Minima To find maximum or minimum points we take the following steps: 1. Find the point at which m = 0 (slope = 0). 2. Tests:

(a) Near point test i. If the slope changes from ( )− to (+), the point is a minimum.

ii. If the slope changes from (+) to ( )− , the point is a maximum. iii. If the slope does not change in sign the point is neither a maximum nor a

minimum. (b) Second derivative test

i. If = (y′′ )− , the point is a maximum ii. If = (+), the point is a minimum y′′

iii. If = 0, test fails, use near point test y′′

If you need to review these tests, refer to module C15 − Applications of Derivatives.

EXAMPLE 1 Find any maximum and/or minimum points for: 23 xy x e= SOLUTION:

2 2

2 2

2

(3 )( )(2) 3

6 33 (2 1)

x x

x x

x

dy x e edx

x e ee x

= +

= +

= +

Setting the derivative to zero and solving for x, 2(3 )(1 2 ) 0xe x+ = Set each factor to zero:

2

2

3 0 2 10 1

x

x

e xe x

= +

= =

0/ 2=

− Since 2xe is always positive, 2xe can never be equal to zero. Therefore only the point

represents a possible maximum or minimum point. 1 2/x = −22 22


Test the point . Finding the second derivative of this function may be difficult, therefore use the near point test.

1 2x = − /

2

2 0

at 1, 3 ( 1) 0 406

at 0, 3 (1 2 ) 3 =3 x

dyx edx

dyx e x edx

−= − = − = − .

= = + =

The slope changes from negative to positive. Therefore the point ( 0 5 0 5518)− . ,− . is a minimum point. Inflection Points From A to B the curve is concave down.

23

From B to C the curve is concave up. The point B represents the inflection point because it is the point where the curve changes its concavity. To find an inflection point: 1. Find the second derivative and equate it to zero. 2. Test the point:

(a) Near point test. i. If changes in passing through the point, this point is then an inflection

point y′′

ii. If does not change, the point is not an inflection point y′′(b) Third derivative test

i. If the value of the third derivative at the point is positive or negative, then the point is an inflection point

ii. If the value of the third derivative is zero, the test fails, then use the near point test

EXAMPLE 2 Find any maximum, minimum and/or inflection points for the following curve:

2 xy x e−= SOLUTION:

2

2

2

( )( 1)( ) 2

2( )(2 )

x

x x

x x

x

y x edy e x xdx

x e xexe x

−

− −

− −

−

=

= − +

= − +

= −

e

A B C


To find the maximum or minimum points, set the derivative to zero. Therefore: ( )(2 ) 0xxe x− − =

0 and (2 ) 00 2

x

x

xe xxe x

−

−

= − =

= =

now setting each factor to zero:

Since Since xe− cannot be equal to zero, we have 2 critical values of x = 0 and x = 2. It is left to the student to test these values to determine whether they are maximum or minimum. Use the near point test. To determine whether the given function has any inflection points, find the second derivative and set it to zero.

( )( ) (

2

2

2 2

2

2 2 2 4 2

x

x

x x

y x e

y x x e

y x x x e x x e

−

−

− −

=

′ = −

′′ = − + + − = − + =) 0 solving the second derivative equals zero 2 4 2 0x x− + = using the quadratic formula 2 2x = ± Again it will be left to the student to test these values.

2 2 and 2

3 4142 0 5858x x= + = −= . = .

2

Since finding the third derivative of the function may not be easy, it is suggested that the near point test be used. x = 3.4142 and x = 0.5858 are the x coordinates of the points of inflection. The following sketch is typical of a class of curves which correspond to the “type” of expression in this example.

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Max Observe here we have 2 points of inflection and 2 max/min points

P.I

Max

P.I

y

x


Experiential Activity Nine 1. Given find any maximum, minimum and/or inflection points. (7 8)4 xy xe −= 2. Given 3 36 xy x e−= find any maximum, minimum and/or inflection points. 3. Given 35 xy x e= find any maximum, minimum and/or inflection points. 4. Given 2 3xy x e−= find any maximum, minimum and/or inflection points. 5. Given 22 xy x e−= find any maximum, minimum and/or inflection points.

Experiential Activity Nine Answers 1. min

inflection point

5( 0 1429 7 0520 10 )−− . , − . × →5( 0 2857 5 1886 10 )−− . , − . × →

2. max

inflection point inflection point

inflection point

(1 0 2987), . →(0 0), →(0 4226 0 1275. , .(1 5774 0 2074). , .

)→→

3. min

inflection point inflection point

inflection point

( 3 6 7213)− , − . →( 4 7321 4 6668− . , − .( 1 2679 2 868− . , − .(0 0), →

) →2) →

4. min (0 0), →

23( 0 0601), . →

(0 1953 0 0212). , .(1 1381 0 0426. , .

max inflection point inflection point

→) →

5. min

max inflection point inflection point

(0 0), →(2 1 082, .(0 5858. ,(3 4142. ,

7) →0 3820).0 767.

→1) →

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Practical Application Activity Complete the Differentiation of Exponential Functions assignment in TLM.

Summary Differentiation of exponential functions was covered in detail. Functions of the form

and in particular uy b= uy e= were differentiated alone and in combination with other expressions. This module prepares the student well, for applied areas especially in the field of electronics.

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Documents

Differentiation of Exponential Functionstlment.nait.ca/.../C33-Differentiation_of_Exponential_Functions.pdf · Module C33 − Differentiation of Exponential Functions EXAMPLE 1 Find