5.13 Derivatives of the Inverse Trigonometric Functions

In the previous section, we learned the relation between the derivative of a function and that of its inverse. Because we know the derivatives of trigonometric functions, in this section we derive the derivatives of the inverse trigonometric functions.

Recall $(f^{- 1})^{'} (y) = \frac{1}{f^{'} (x)} = \frac{1}{f^{'} (f^{- 1} (y))}$

Table of Contents

Derivative of the Inverse of Sine

Recall:

$v = \arcsin u$ means $u = \sin v$ and $- \frac{π}{2} \leq v \leq \frac{π}{2}$ .

We can show that

$\frac{d}{d x} \arcsin x = \frac{d}{d x} \sin^{- 1} x = \frac{1}{\sqrt{1 - x^{2}}}, (- 1 < x < 1) .$

Figure 1: The graphs of

y = \arcsin x

and

y = \sin x

Example 1

Find $\frac{d}{d x} \arcsin x$

Solution

Let $y = \sin x$ . Then $x = \arcsin y$ and $- π / 2 \leq x \leq π / 2$

$\begin{aligned} \frac{d}{d y} \arcsin y & = \frac{1}{\frac{d}{d x} \sin x} \\ = \frac{1}{\cos x} \\ = \frac{1}{\sqrt{1 - \sin^{2} x}} & (\cos x \geq 0 when \frac{- π}{2} \leq x \leq \frac{π}{2}) \\ = \frac{1}{\sqrt{1 - y^{2}}} & (\sin x = y) \end{aligned}$

With a change in notation we can express the above result as
$\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}} .$

Example 2

Find $\frac{d}{d x} (\arcsin x^{2}) .$

Solution

Letting $u = x^{2}$

$\begin{aligned} \frac{d}{d x} \arcsin x^{2} & = \frac{d}{d u} \arcsin u \cdot \frac{d}{d x} u \\ = \frac{1}{\sqrt{1 - u^{2}}} (2 x) \\ = \frac{2 x}{\sqrt{1 - x^{4}}} . \end{aligned}$

Derivative of the Inverse of Cosine

Recall

$v = \arccos u = \cos^{- 1} u$ means $u = \cos v$ and $0 \leq v \leq π$ .

We can show that

$\frac{d}{d x} \arccos x = \frac{d}{d x} \cos^{- 1} x = - \frac{1}{\sqrt{1 - x^{2}}}, (- 1 < x < 1) .$

Figure 2: The graphs of

y = \arccos x

and

y = \cos x

Example 3

Find $\frac{d}{d x} \arccos x or \frac{d}{d x} \cos^{- 1} x$

Solution

If $y = \cos x$ , then $x = \arccos y$ and
$0 \leq x \leq π$
$\begin{aligned} \frac{d}{d y} \arccos y & = \frac{1}{\frac{d}{d x} \cos x} \\ = \frac{1}{- \sin x} \\ (\sin x \geq 0 when 0 \leq x \leq π) & = \frac{- 1}{\sqrt{1 - \cos^{2} x}} \\ = \frac{- 1}{\sqrt{1 - y^{2}}} \end{aligned}$
With a change in notation we can express the above result as
$\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}} .$

Derivative of the Inverse of Tangent

Recall

$v = \arctan u = \tan^{- 1} u$ means $u = \tan v$ and $- \frac{π}{2} \leq v \leq \frac{π}{2}$ .

We can show that
$\frac{d}{d x} \arctan x = \frac{d}{d x} \tan^{- 1} x = \frac{1}{1 + x^{2}} .$

Figure 3: The graphs of

y = \arctan x

and

y = \tan x

Example 4

Find

\frac{d}{d x} \arctan x or \frac{d}{d x} \tan^{- 1} x

Solution

If $y = \tan x$ , then $x = \arctan y$ and $- π / 2 \leq x \leq π / 2$
$\begin{aligned} \frac{d}{d y} \arctan y & = \frac{1}{\frac{d}{d x} \tan x} \\ = \frac{1}{1 + \tan^{2} x} \\ = \frac{1}{1 + y^{2}} \end{aligned}$
With a change in notation, we get $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}} .$