Derivatives of the Inverse Trigonometric Functions

Section 3–6: 

The Derivatives of the Inverse Trigonometric Functions 

If y = arcsin(u) then 

If y = arccos(u) then 

If y = arctan(u) then 

dy 

dx = 

du 

1− u 2 

dy 

dx = 

−du 

1− u 2 

dy 

dx = 

du 

1 + u 2 

If y = arc csc(u) then 

If y = arc sec(u) then 

If y = arc cot(u) then 

dy 

dx = 

u 

− du 

u 2 −1 

dy 

dx = 

u 

du 

u 2 −1 

dy 

dx = −du 

1 + u 2 

The Derivative of the Inverse Sine Function 

The Inverse Sine Function is written as y = arcsin(x) or y = sin −1 (x) 

If y = arcsin(u) then 

dy 

dx = 

du 

1− u 2 

Example 1 Example 2 Example 3 

Find dy 

dx 

if 

( ) 

y = arcsin 3x 2 

u = 3x 2 

y ′ = 

du = 6x 

6x 

( ) 2 

1− 3x 2 

Find dy if 

dx 

y = arcsin 2x − 5 

u = 2x − 5 du = 2 

y ′ = 

Find dy 

dx 

( ) 

2 

( ) 2 y ′ = 

1− 2x − 5 

if 

( ) 

y = sin −1 tanx 

u = tanx du = sec 2 x 

sec 2 x 

1− ( tanx) 2 

y ′ = 

6x 

1− 9x 4 

y ′ = 

sec 2 x 

1− tan 2 x 

Math 400 3–6 Page 1 of 8 ©2012 Eitel

The Derivative of the Inverse Cosine Function 

The Inverse Cosine Function is written as 

y = arccos(x) or y = cos −1 (x) 

If y = arccos(u) then 

dy 

dx = 

−du 

1− u 2 


Find dy 

dx 

if 

( ) 

y = arccos 4x 2 

Find dy 

dx 

if 

( ) 

y = arccos e x 

Find dy if 

dx 

y = arccos sin x 

( ) 

u = 4x 2 

du = 8x 

u = e x 

du = e x 

u = sinx du = cos x 

y ′ = 

y ′ = 

−8x 

( ) 2 

1− 4 x 2 

−8x 

1−16x 4 

y ′ = 

y ′ = 

−e x 

( ) 2 

1− e x 

−e x 

1− e 2x 

y ′ = 

y ′ = 

−cos x 

1− sin 2 x 

−cos x 

cos 2 x 

y ′ = 

−cosx 

cosx 


The Derivative of the Inverse Tangent Function 

The Inverse Tangent Function is written as y = arctan(x) or y = tan −1 (x) 


dy 

dx = 

du 

1 + u 2 


Find dy 

Find dy if 

if 

dx 

dx 

y = arctan( x ) 

y = arctan( 3x 2 ) 

u = x du = 1 

2 x 

u = 3x 2 du = 6x 

1 

6x 

y ′ = 

1 + ( 3x 2 ) 2 

y ′ = 

2 x 

1 + ( x ) 2 

6x 

1 

y ′ = 

1 + 9x 4 y ′ = 

2 x 

1 + x 

Find dy 

dx 

if 

( ) 

y = tan −1 sin x 

u = sinx du = cos x 

y ′ = 

cosx 

1 + sin 2 x 

y ′ = 

1 

2 x 1 + x ( ) 


The relationship between the The relationship between the The relationship between the 

arcsec and arc cos functions arccsc and arcsin functions arctan and arccot functions 

⎛ 1⎞ 

arcsec (x) = arccos 

⎝ x⎠ 

⎛ 1⎞ 

arccsc (x) = arcsin 

⎝ x⎠ 

⎛ 1⎞ 

arccot (x) = arctan if x > 0 

⎝ x⎠ 

⎛ 1⎞ 

arccot (x) = π + arctan if x < 0 

⎝ x⎠ 

If you wanted to find the derivative of an arcsec function, you could convert it to an arccos function using the 

rules listed above and then take the derivative of the arccos function. This same technique could be used 

to find the derivatives for arccsc and arccot functions. In most cases we do not do this. We use the 

derivative rules listed below and take the derivatives directly. 




dy 

dx = 

u 

du 

u 2 −1 

dy 

dx = 

u 

− du 

u 2 −1 

dy 

dx = −du 

1 + u 2 

The Derivative of the Inverse Secant Function y = arcsec (x) or y = arcsec −1 (x) 


dy 

dx = 

u 

du 

u 2 −1 

Example 1 Example 2 

Find dy 

dx 

if 

( ) 

y = arcsec 3x 2 

Find dy 

dx 

if 

( ) 

y = arcsec e x 

u = 3x 2 

du = 6x 

u = e x 

du = e x 

y ′ = 

6x 

3x 2 3x 2 

( ) 2 −1 

y ′ = 

e x 

e x 

( e x ) 2 −1 

y ′ = 

3x 2 

6x 

9x 4 −1 

y ′ = 

e x 

e x 

( e x ) 2 −1 

y ′ = 

x 

2 

9x 4 −1 

y ′ = 

1 

e 2x −1 


The Derivative of the Inverse Cosecant Function y = arccsc (x) or y = arccsc −1 (x) 


dy 

dx = 

u 

− du 

u 2 −1 

Example 1 Example 2 

Find dy 

dx 

if 

( ) 

y = arccsc x 2 − 2 

Find dy 

dx 

if 

( ) 

y = arccsc 2x 3 

u = 

x 2 − 2 du = 2x 

u = 2x 3 du = 6x 2 

y ′ = 

−2x 

x 2 − 2 ( x 2 − 2) 2 −1 

y ′ = 

−6x 2 

2x 3 2x 3 

( ) 2 −1 

y ′ = 

2x 3 

−6x 2 

4x 6 −1 

y ′ = 

2x 2 x 

−6x 2 

4 x 6 −1 

y ′ = 

x 

−3 

4x 6 −1 


Derivatives with Inverse Trigonometric Functions and Polynomial functions 

Example 1 

Quotient Rule 

Find dy 

dx 

if y = arcsin ( x) 

x 

der 

bottom 67 8 64 7 

of 

48 

top 

1 

y ′ = x • 

1− x 2 

top der 

64 748 

6 of 7 bottom 8 

− arcsin x • 1 

( ) 

y ′ = x • 

1 

1− x 2 

− arcsin ( x) 

•1 

y ′ = 

x 

( ) 

1− x 2 − arcsin x 

Example 2 

Product Rule 

Find dy 

dx 

if y = x 2 arctan( x) 

first der 

67 8 64 7 of 4 sec 8 

y ′ = x 1 1 

• 

1− x 2 

64 second 748 

6 der 47 

of first 48 

+ arcsin x • 2x 

( ) 

y ′ = 

x 2 1 

• 

1 + x 2 + arctan x 

= x 2 

+ 2x arctan ( x ) 

1 + x 

2 

( ) •2x 


Example 3 

Power Rule 

Find dy 

dx 

if y = 

( arctan( x) 

) 2 

n} 64748 

y ′ = 2 • arctan x 

u n−1 

( ( )) 1 

67 du8 

1 

• 

1 + x 2 

( ( ) 1 • 

y ′ = 2 • arctan x 

y ′ = 

2arctan( x) 

1 + x 2 

1 

1 + x 2 

Example 4 

Find dy 

dx 

if y = arcsin ( x) + arccos ( x) 

y ′ = 

1 

1− x 2 + 

−1 

1− x 2 

y ′ = 0 


Find dy 

dx 

Example 5 

if y = arcsin ( x) + x • 1− x 2 

( ) 1/2 

y = arcsin( x) + x • 1− x 2 

dy 

dx = 

d dx arcsin ( x) 

+ d 

dx 

( ) 1/2 

⎛ 

⎜ x • 1− x 2 

⎝ 

⎞ 

⎟ 

⎠ 

y ′ = 

der 

6 4 

of 

7 

arcsin(x) 

4 8 

1 

1− x 2 

6 first 78 

6 

+ x • 1 44 der 7 of 4 sec 448 

( 1− x2 ) −1/2 •−2x 

2 

64 second 74 

8 der 67 of 8 first 

• + 1− x 2 • 1 

y ′ = 

1 ⎡ 

1− x 2 + x • 1 2 1− x 2 

⎣ 

⎢ 

( ) −1/2 • −2x 

⎤ 

⎦ 

⎥ + 

1− x2 

y ′ = 

1 

1− x 2 − x 2 

1− x 2 + 1− x 2 

add the first two terms with a common denominator 

( ) 

y ′ = 1− x 2 

1− x 2 + 

y ′ = 

( 1− x2 ) 1 

( 1− x 2 ) 

1− x2 

1/2 

+ 1− x 

2 

y ′ = 1− x 2 + 1− x 2 

y ′ = 2 1− x 2

Derivatives of the Inverse Trigonometric Functions

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