Calculus AB Chapter 3.6 – 3.9 Exam Review Solutions ****Here's a quick list of answers … explanations follow …****

1. 15

2. v (t) = 5t ( ) 25

2 5s t t= +

3. a) 4

2'1

xyx

=−

b) 2

cos'1 sin

xyx

=+

c) 121

2 11 2 1

xx x x x

−

=− −

d) 2

2xe x− ⋅

e) ( )2 55 ln 5 2x x+ ⋅ ⋅

f) ( ) ( ) ( )2 2 1

2 2

1 4 4 ln 4 1 4 sin 2 ln 4yx x −=

− ⋅ ⋅ − ⋅ ⋅x

g) ( ) ( )' cot ln sin sinxexy e x x x⎡ ⎤= + ⋅⎣ ⎦

4. a) ( )sin' 3 ln 3 cosxy x= ⋅ ⋅

b) y ' = 1

5. 2

1 26

dy xydx x y

−=

+

6. ( ) ( )4 ''' 13440 4 1y x x= +

****Explanations/Work**** 1. Since the derivative of the inverse function is the reciprocal of the derivative of the function at the inverse point, we must find what values of x give 2 for the original function. In other words,

( ) ( )( )

'1 12' ?

ff

− = , and we need to find the value of "?".

Solve the equation 32 2 1x x= + − for x by setting the equation equal to zero, and using your calculator to solve for the zeros. You get x = 1. Since , we get ( ) 2' 3f x x= +2

( ) ( )( ) ( )

'12

1 12' 1 53 1 2

ff

− = = =+

1

2. Since a (t) = 5, then v (t) = 5t + C, where C is some constant. Use the fact that v (2) = 10, to get 10 = 5(2) + C, which means C = 0 and v (t) = 5t. Since v (t) = 5t, then ( ) 25

2s t t= +K , where K is some constant.

Use the fact that s (0) = 5, to get ( )2525 0 K= + , which means K = 5 and ( ) 25

2 5s t t= + .

3. a) Using the rule for the derivative of inverse sine … 4

2'1

xyx

=−

.

b) Using the rule for the derivative of inverse tangent … 2

cos'1 sin

xyx

=+

c) Using the rule for the derivative of inverse secant … 121

2

1x

x x

−

−. Writing this as one fraction with positive exponents

gives us 1

2 1x x x−

Since 2x x x= = x , we could write the answer as

12 1x x−

.

d) Using the rule for the derivative of … ue

2

2xe x− ⋅ e) Using the rule (where a is a constant and u is a differentiable function) … ( )ln 'u ua a a u= ⋅ ⋅ ( )

2 55 ln 5 2x x+ ⋅ ⋅ f) Using the rules of logarithms, we can rewrite this to be

( )14 sin 2y x−= . Using implicit differentiation, we have

( )2

24 ln 41 4

y dydx x

⋅ ⋅ =−

Solving for dydx

, we get

( )2

2

1 4 4 ln 4yx− ⋅ ⋅

Since ( )14 sin 2y x−= , we could write

( ) ( )2 1

2

1 4 sin 2 ln 4x x−− ⋅ ⋅

g) Using logarithmic differentiation, we can turn ( )sin

xey x= into

( )ln ln sinxy e x= ⋅ Using implicit differentiation, we have

( )' cos ln sinsin

x xy xe xy x= ⋅ + ⋅e .

Factoring out an xe on the right side, and multiplying both sides by y, we have

( )cos' ln sinsin

x xy e x yx

⎡ ⎤⎢ ⎥= + ⋅⎢ ⎥⎣ ⎦

Since ( )sinxey x= , we can write

( ) ( )' cot ln sin sinxexy e x x x⎡ ⎤= + ⋅⎣ ⎦

4. a) Using the rule (where a is a constant and u is a differentiable function) … ( )ln 'u ua a a u= ⋅ ⋅ ( )sin3 ln 3 cosx x⋅ ⋅ b) Since the original function can be simplified to y = x as long as x > 0 … y ' = 1 . 5. Using implicit differentiation and simplifying we have,

( )

2

2

2

2

2 6 1

6 1 2

6 1 2

1 26

dy dyx y x ydx dx

dy dyx y xdx dx

dy

y

x y xdxdy xydx

y

x y

⎡ ⎤⎢ ⎥+ ⋅ + =⎢ ⎥⎣ ⎦

+ = −

+ = −

−=

+

6. Using the chain rule (a few times) … ( )

( )

( )

( )

( )

( )

6

6

5

5

4

4

' 7 4 1 4

28 4 1

'' 168 4 1 4

672 4 1

''' 3360 4 1 4

13440 4 1

y x

x

y x

x

y x

x

= + ⋅

= +

= + ⋅

= +

= +

= +

⋅