5 Steps to a 5 AP Calculus BC 2019

84 STEP 4. Review the Knowledge You Need to Score High

Example 8

Ify=

tanx

1 +tanx

, find

dy

dx

.

Using the quotient rule, letu=tanxandv=(1+tanx). Then,

dy

dx

=

(sec^2 x)(1+tanx)−(sec^2 x)(tanx)

(1+tanx)^2

=

sec^2 x+(sec^2 x)(tanx)−(sec^2 x)(tanx)

(1+tanx)^2

=

sec^2 x

(1+tanx)^2

, which is equivalent to

1

(cosx)^2

1 +

(

sinx

cosx

) 2

=

1

(cosx)^2

(

cosx+sinx

cosx

) 2 =

1

(cosx+sinx)^2

.

Note: For all of the above exercises, you can find the derivatives by using a calculator,

provided that you are permitted to do so.

Derivatives of Inverse Trigonometric Functions

Summary of Derivatives of Inverse Trigonometric Functions

Letube a differentiable function ofx, then

d

dx

sin−^1 u=

1

√

1 −u^2

du

dx

, |u|< 1

d

dx

cos−^1 u=

− 1

√

1 −u^2

du

dx

,|u|< 1

d

dx

tan−^1 u=

1

1 +u^2

du

dx

d

dx

cot−^1 u=

− 1

1 +u^2

du

dx

d

dx

sec−^1 u=

1

|u|

√

u^2 − 1

du

dx

, |u|> 1

d

dx

csc−^1 u=

− 1

|u|

√

u^2 − 1

du

dx

, |u|> 1.

Note that the derivatives of cos−^1 x, cot−^1 x, and csc−^1 xall have a “−1” in their numerators.

Example 1

Ify=5 sin−^1 (3x), find

dy

dx

.

Letu= 3 x. Then

dy

dx

=(5)

1

√

1 −(3x)^2

du

dx

=

5

√

1 −(3x)^2

(3)=

15

√

1 − 9 x^2

.

Or using a calculator, enterd[5 sin−^1 (3x), x] and obtain the same result.