Section 4.2 – Differentiating Exponential Functions

x

x x

x

f ' x e f '

f x e f x

x l

a

na a

THE MEMORIZATION LIST BEGINS

Find dy

:dx

x 1y e

x 1dye

dx

x 1y 2

x 1dyln2 2

dx

2 xy 2x e 3

xdy4x e

dx

2 / 3 xy 3x 7

1/ 3 xdy2x ln7 7

dx

x x 3y 4 e 3x

x x 2dyln4 4 e 9x

dx

POSITION s(t)

VELOCITY v(t)

ACCELERATION a(t)

DIFFERENTIATE

INTEGRATE

A particle moves along a line so that at time t, 0 < t < 5, its position

is given by t t 2s t 5 2e t 1

a) Find the position of the particle at t = 2

2 2 2s 2 5 2e 2 1 229 2e

b) What is the initial velocity? (Hint: velocity at t = 0)

t ts' t v t ln5 5 2e 2t

0 0v 0 ln5 5 2e 2 0 ln5 2

c) What is the acceleration of the particle at t = 2

2 t tv ' t a t ln5 5 2e 2

2 2 2a 2 ln5 5 2e 2

CALCULATOR REQUIRED

Suppose a particle is moving along a coordinate line and its position at time t is given by

2

2

9ts t

t 2

For what value of t in the interval [1, 4] is the instantaneousvelocity equal to the average velocity?

a) 2.00 b) 2.11 c) 2.22 d) 2.33 e) 2.44

ave

f 4 f 1 5v

4 1 3

If f x F S F'S 'f ' x S F

NIf f x

D then N'D

f ' x 2D

D'N

NIf f x

D

then

N'DIf f ' x

2D D

D'N

The Quotient Rule

An equation of the normal to the graph of xf x at 1,f 1 is

2x 3

NO CALCULATOR

A) 3x y 4

B) 3x y 2

C) x 3y 2

D) x 3y 4

E) x 3y 2

1

f 1 12 1 3

2

1 2x 3 2 xf ' x

2x 3

2

1 2 1 3 2 1 3f ' 1 3

12 1 3

1y 1 x 1

3

3y 3 x 1

2

2 2 2 22 2 2 2

3 dyIf y ,

4 x dx3 3x 6x 6x 3

A) B) C) D) E)2x 1 x 4 x 4 x 4 x

NO CALCULATOR

2

22

0 4 x 2x 3dy

dx 4 x

3If g x x 1 and f is the inverse function of g, then f ' x

2 4 / 3 2 / 32 1 1A) 3x B) 3 x 1 C) x 1 D) x 1 E) DNE

3 3

3x y 1 3x y 1 3y x 1

2dy3x

dx

NO CALCULATOR

An equation of the line normal to the curve xf x at 1, 1 is :

x 2

NO CALCULATOR

2

1 x 2 1 xf ' x

x 2

2

1 1 2 1 1f ' 1 2

1 2

1y 1 x 1

2

A) 2x y 3 0

B) 2x y 1 0

C) x 2y 3 0

D) x 2y 1 0

E) x 2y 3 0

2y 2 x 1

2 2

x kIf f x and k 0, then f " 0

x k4 2 2 4

A) B) C) 0 D) E)k k k k

NO CALCULATOR

2 2

1 x k 1 x k 2kf ' x

x k x k

2

4

0 x k 2 x k 2kf " x

x k

2

4 4 2

2 0 k 2k 4k 4f " 0

k k0 k

x 1ln eLet f x for x 0

2xIf g is the inverse of f, then g' 1

A) 2 B) 1 C) 0 D)1 E) 2

NO CALCULATOR

x 1f x

2x

g x 1x

2g x

2xg x g x 1

g x 2x 1 1

1g x

2x 1

2

0 2x 1 2 1g' x

2x 1

2

0 2 1 1 2 1g' 1

2 1 1

Consider the function 3

6xf x where f ' 0 3. Then a

a x

A) 5 B) 4 C) 3 D) 2 E) 1

3 2

23

6 a x 3x 6xf ' x

a x

23

23

6 a 0 3 0 6 0f ' 0 3

a 0

6

3a

NO CALCULATOR

If u(4) = 3, u ‘ (4) = 2, v(4) = 1, v ‘ (4) = 4, find:

x 4

d u|

dx v

x 4

duv |

dx

x 4

d2u 3v |

dx

x 4

d 2v|

dx u

2

u' 4 v 4 v ' 4 u 4

v 4

2

2 1 4 310

1

u' 4 v 4 v ' 4 u 4 2 1 4 3 14

2u' 4 3v ' 4 2 2 3 4 8

2

2v ' 4 u 4 2u' 4 v 4

u 4

2 4 3 2 2 1 20

9 9

ADD TO THE MEMORIZATION LIST

ddx

xsinx

cos ddx

nco x

xs

si

2ddx

san

xx

ect

2ddx

cot xcsc x

sd ese

c xdx

c xtanx csc x cot xd

dx

csc x

4.1

cosxf x 2 3sinx

sinxf ' x 2 osx3c

sin xf x x cos

cosx sincosx s xf inx' x

2 2f ' x cos x sin x

f ' x cos2x

f xsinx

x

2

xf ' x

cosx 1 sinx

x

2

x cosx sinxf ' x

x

2x x xf cos

2cosx2xf x x' sinx

2f ' x 2x cosx x sinx

sec xf x x tan

2tanxsec x tanx secf ' c xx x se

2 3f ' x sec x tan x sec x

2 3f ' x sec x sec x 1 sec x

3f ' x 2sec x sec x

sec xf x

tanx

2

2

tanxsec x tanx se sec xc

t

x'

xf x

an

2 3

2

sec x tan x sec xf ' x

tan x

sec xf x

tanx sec x cot x 2cot xs sece xc x tanf x csc' xx

2f ' x sec x sec x csc x

2

3 2

1 cos xsec x

cos x sin x

2sec x sec x csc x 2sec x 1 csc x 2sec x cot x

2sec x 1 csc x 2sec x cot x

2

2

1 cos x

cos x sin x

cos x 1

sinx sinx

csc xcot x

2

2

1 cos x

cos x sin x

cos x 1

sinx sinx

csc xcot x

sec xf x

tanx 1 cos x

cos x sinx csc x f ' x csc xcot x

1f x

cot x

2

2

0 cscf ' x

xcot x 1

cot x

2

2

csc xf ' x

cot x

2

2 2

1 sin x

sin x cos x

2

1

cos x 2sec x

1f x

cot x

2f ' x sec x

tanx

f x x cosx

f ' x 1 sinx

f x csc x sinx

f ' x csc x cot x cosx

2

2 2

cosx sin x cosxf ' x

sin x sin x

22

cosxf ' x 1 sin x

sin x

2f ' x cosx cot x

3 1 sinxf x

2cosx

3 3f x sec x tanx

2 2

23 3f ' x sec x tanx sec x

2 2

3f ' x sec x tanx sec x

2

2f x 2x sinx x cosx

2f ' x 2sinx 2x cosx 2x cosx x sinx

2f ' x 4x cosx 2 x sinx

Find the equation of the tangent line of f x cos x at x2

ff ' xa ax f a

f cos 02 2

ff 02

x x'2

x' x nf si

sin2

f ' 12

x x10f2

NO CALCULATOR

At x = 0, which of the following is true of x

1f x sinx ?

e

A) f is increasingB) f is decreasingC) f is discontinuousD) f is concave upE) f is concave down

x x

2x

0 e e 1f ' x cos x

e

x

1f ' x cos x

e 0

1f ' 0 cos0 0

e

XX

0

1f 0 sin0 1

e

X

x x

2x

0 e e 1f " x sinx

e

x

1f " x sinx

e

0

1f " 0 sin0 1

e

NO CALCULATOR

If the average rate of change of a function f over the intervalfrom x = 2 to x = 2 + h is given by h7e 4cos 2h , then f ' 2

A) -1 B) 0 C) 1 D) 2 E) 3

hf 2 h f 2

7e 4cos 2h2 h 2

h 0

h

h 0

f 2 h f 27e 4cos 2h

2 hf ' 2 lim l

2im

7 1 4cos 0

7 4

NO CALCULATORThe graph of f x x sinx defined on 0 x has an inflection point whenever

2 2A) tanx B) tanx C) tanx x D) sinx x E) cos x x

x x

f ' x 1sinx xcos x

f " x cos x cos x sinx x

0 2cos x sinx x

x sinx 2cos x

2tanx

x

NO CALCULATOR

h 0

tan x h tanxlim

h

2 2A) sec x B) sec x C) sec x D) sec x E) DNE

3If F x 3x sinx cos x

2

NO CALCULATOR

then an equation of the line tangent

to the graph of F at the point where x is2

A) y 3x B) y 3 x C) y 4x D) y 4x E) y 4x 22 2 2

3F x 3x sinx cos x

2

3

3 sin cos2 2 2

F 022

F' x 3sinx 3xcos x sinx

3sin 3 cos sin2 2 2 2

F' 42

x2

0y 4