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

We can show that

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

Figure 1: The graphs of $y=\arcsin x$ and $y=\sin x$.

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

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

Figure 2: The graphs of $y=\arccos x$ and $y=\cos x$

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

Figure 3: The graphs of $y=\arctan x$ and $y=\tan x$.

Example 4

Find\[\frac{d}{dx}\arctan x\quad\text{ or }{\quad\frac{d}{dx}\tan^{-1}x}\]

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 \[\frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}.\]