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 \[(f^{-1})'(y)=\frac{1}{f'(x)}=\frac{1}{f'(f^{-1}(y))}\]

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

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

[fig:Trig-ArcSin-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
\(-\pi/2\leq x\leq\pi/2\)
\[\begin{align} \dfrac{d}{dy}\arcsin y & =\frac{1}{\dfrac{d}{dx}\sin x}\\ & =\frac{1}{\cos x}\\ & =\frac{1}{\sqrt{1-\sin^{2}x}}\tag{${\cos x\geq0}\text{ when }{\frac{-\pi}{2}\leq{x}\leq\frac{\pi}{2}}$}\\ & =\frac{1}{\sqrt{1-y^{2}}}\tag{${\sin x=y}$}\end{align}\]
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

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

[fig:ch5-trig-ArcCos]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
\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}\arctan x=\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}.}\]

[fig:ch5-trig-ArcTan]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 \[\frac{d}{dx}\arctan x=\frac{1}{1+x^{2}}.\]

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