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Differentiate y = ( 3 x 2 + 8e x )

1/ 3

using chain rule

1
dy 1
2
x 3 −1 d
= ( 3 x + 8e )
3 x 2 + 8e x )
(
dx 3
dx
−2 / 3
1
= ( 3 x 2 + 8e x ) ( 6 x + 8e x )
3

=

6 x + 8e x
3 ( 3 x 2 + 8e x )

2/3

Differentiate f ( x ) = sin 4 x using chain rule
f ' ( x ) = 4sin 4−1 x

d
sin x
dx

= 4sin 3 x cos x
f ' ( 2 ) = 4sin 3 2 cos 2

Differentiate f ( x ) = 7 x 2 − 8e x using chain rule
d
( 7 x 2 − 8e x )
dx
d
d
= 7 x2 − 8 ex
dx
dx
x
= 7  2 x − 8e

f '( x) =

= 14 x − 8e x
d
f '' ( x ) = (14 x − 8e x )
dx
= 14 − 8e x
f '' ( 4 ) = 14 − 8e 4

Differentiate f ( x ) = ( −8 x 5 + 10 x8 )( 4e x + 3) using product rule
d
( −8 x5 + 10 x8 )( 4e x + 3) 

dx 
d
d
= ( −8 x5 + 10 x8 ) ( 4e x + 3) + ( 4e x + 3) ( −8 x 5 + 10 x8 )
dx
dx
= ( −8 x5 + 10 x8 )( 4e x + 0 ) + ( 4e x + 3)( −8  5 x 4 + 10  8 x 7 )

f '( x) =

= ( −8 x5 + 10 x8 ) 4e x + ( 4e x + 3) ( −40 x 4 + 80 x 7 )

ex
using quotient rule rule
2− x
d
d
( 2 − x ) ex − ex ( 2 − x )
dx
dx
f '( x) =
2
(2 − x)

Differentiate f ( x ) =

( 2 − x ) e x − e x ( 0 − 1)
2
(2 − x)
(3 − x ) ex
=
2
(2 − x)

=

Differentiate f ( x ) = ( 5 x + 3)
f '( x) =

d
−1
( 5 x + 3)
dx

= −1( 5 x + 3)

−1−1

= −5 ( 5 x + 3)
f ' ( 4 ) = −5 ( 20 + 3)
= −

5
529

−2

−2

(5)

−1

using chain rule

Differentiate s ( t ) = 6t 3 + 18t 2 − 54t to get v ( t ) .
v (t ) = s ' (t )
d
( 6t 3 + 18t 2 − 54t )
dt
= 6  3t 2 + 18  2t − 54

=

= 18t 2 + 36t − 54
= 18 ( t 2 + 2t − 3)
= 18 ( t + 3)( t − 1)
The velocity becomes zero at :
t = −3 and 1
a (t ) = v '(t )
d
18 ( t 2 + 2t − 3)
dt
= 18 ( 2t + 2 − 0 )

=

= 36 ( t + 1)

Differentiate f ( x ) = cos ( x ) to get
f ' ( x ) = − sin x
f '' ( x ) = − cos x
f ''' ( x ) = sin x
f ( 4) ( x ) = cos x
34 = 4  8 + 2
d d

 cos x 
dx  dx

d
= ( − sin x )
dx
= − cos x
So, the 34th derivative of cos x is − cos x.
f ( 34) ( x ) =

The answer is − cos x

Differentiate s = t 5 − 6t 4 to get
d
s ' ( t ) = ( t 5 − 6t 4 )
dt
= 5t 4 − 6  4t 3
= 5t 4 − 24t 3
a ( t ) = s '' ( t )
d
( 5t 4 − 24t 3 )
dt
= 5  4t 3 − 24  3t 2

=

= 4t 2 ( 5t − 18 )
a (t ) = 0
4t 2 ( 5t − 18 ) = 0
18
5
18
t=
5

t = 0,

s ( t ) = 7 − 3sin ( t )
Initial height = 7 inches
v ( t ) = s ' ( t ) = 0 − 3cos ( t ) = −3cos ( t )
v ( / 2 ) = −3cos ( / 2 ) = 0
The velocity is zero at
t =  / 2 seconds
d
a ( t ) = v ' ( t ) = ( −3cos ( t ) )
dt
in
a ( t ) = 3sin ( t ) 2
s

Differentiate f ( x ) = 6 4 x8 + 5 x 9 using chain rule
1/ 2
d
6 ( 4 x8 + 5 x 9 )
dx
1
−1 d
1
= 6  ( 4 x8 + 5 x 9 ) 2
4 x8 + 5 x 9 )
(
2
dx

f '( x) =

1
9 −2

= 3( 4x + 5x )

( 4  8x + 5  9x )
3 ( 32 x + 45 x )
=
8

7

4 x8 + 5 x 9

8

7

8

Differentiate f ( x ) = −9sin 2 ( −6 x 5 ) using chain rule
d
 −9sin 2 ( −6 x 5 ) 

dx 
d
= −9 sin 2 ( −6 x 5 )
dx
d
= −...