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 $$-\dfrac{\pi}{2}\leq v\leq\dfrac{\pi}{2}$$.

We can show that

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\arcsin x=\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}},\quad(-1<x<1).}$

Example 1

Find $$\dfrac{d}{dx}\arcsin x$$

Solution

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

\begin{align} \dfrac{d}{dy}\arcsin y & =\frac{1}{\dfrac{d}{dx}\sin x}\\ & =\frac{1}{\cos x}\\ & =\frac{1}{\sqrt{1-\sin^{2}x}} &{\small (\cos x\geq 0 \text{ when } \frac{-\pi}{2}\leq{x}\leq\frac{\pi}{2})}\\ & =\frac{1}{\sqrt{1-y^{2}}} & {\small (\sin x=y)} \end{align}

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

Example 2

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

Solution

Letting $$u=x^{2}$$

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

## Derivative of the Inverse of Cosine

Recall

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

We can show that

$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\arccos x=\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^{2}}},\quad(-1<x<1).}$

Example 3

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

Solution

If $$y=\cos x$$, then $$x=\arccos y$$ and
$$0\leq x\leq\pi$$
\begin{align} \frac{d}{dy}\arccos y & =\frac{1}{\frac{d}{dx}\cos x}\\ & =\frac{1}{-\sin x}\\ & =\frac{-1}{\sqrt{1-\cos^{2}x}}\tag{{\sin x\geq0\text{ when }0\leq x\leq\pi}}\\ & =\frac{-1}{\sqrt{1-y^{2}}}\end{align}
With a change in notation we can express the above result as
$\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^{2}}}.$

## Derivative of the Inverse of Tangent

Recall

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

We can show that
$\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\arctan x=\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}.}$

Solution

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