5 Steps to a 5 AP Calculus BC 2019

Differentiation 83

Example 2

Findf′(x)iff(x)=cot(4x−6).
Using the chain rule, letu= 4 x−6. Then f′(x)=[−csc^2 (4x−6)][4]=−4 csc^2 (4x−6).

Or using your calculator, enterd(1/tan(4x−6), x) and obtain

− 4

sin^2 (4x−6)

, which is an

equivalent form.

Example 3

Findf′(x)iff(x)=8 sin(x^2 ).
Using the chain rule, letu=x^2. Thenf′(x)=[8 cos(x^2 )][2x]= 16 xcos(x^2 ).

Example 4

Ify=sinxcos(2x), find
dy
dx

.

Using the product rule, letu=sinxandv=cos(2x).

Then
dy
dx
=cosxcos(2x)+−sin(2x)(sinx)=cosxcos(2x)−2 sinxsin(2x).

Example 5

Ify=sin[cos(2x)], find
dy
dx

.

Using the chain rule, letu=cos(2x). Then

dy

dx

=

dy

du

·

du

dx

=cos[cos(2x)]

d

dx

[cos(2x)].

To evaluate
d
dx
[cos(2x)], use the chain rule again by making another u-

substitution, this time for 2x. Thus,
d
dx
[cos(2x)]=[−sin(2x)]2=−2 sin(2x). Therefore,
dy
dx
cos[cos(2x)](−2 sin(2x))=−2 sin(2x) cos[cos(2x)].

Example 6

Findf′(x)iff(x)= 5 xcscx.
Using the product rule, letu= 5 xandv=cscx. Then f′(x)=5 cscx+(−cscxcotx)
(5x)=5 cscx− 5 x(cscx)(cotx).

Example 7

Ify=

√

sinx, find

dy

dx

.

Rewritey=

√

sinx as y =(sinx)^1 /^2. Using the chain rule, letu=sinx. Thus,

dy

dx

=

1

2

(sinx)−

(^12)
(cosx)=
cosx
2(sinx)

12 =

cosx

2

√

sinx

.