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Derivatives of Inverse Sine and Cosine Functions

Let's consider the inverse trigonometric function \(y=\sin ^{-1}u\). We can isolate for \(u\) and get \(\sin y=u\) for the values \(-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}\) for it to qualify as a function.

Now let's take the derivative of \(\sin y=u\) in respect to \(x\).

\[\frac{du}{dx}=\cos y\frac{dy}{dx}\]

Solving for \(\frac{dy}{dx}\) we get,

\[\frac{dy}{dx}=\frac{1}{\cos y}\frac{du}{dx}\]

Now from the identify \(\cos ^2y+\sin ^2y=1\), isolating from \(\cos y\), we get \(\cos y=\sqrt{1-\sin ^2y}\). Thus, the above equation becomes

\[\frac{dy}{dx}=\frac{1}{\sqrt{1-\sin ^2y}}\frac{du}{dx}\]

From our original equation \(u=\sin y\), we now have an equation with \(u\)


Also, \(y=\sin ^{-1}u\). Thus, the derivative of the inverse sine function is

\[\frac{d(\sin ^{-1}u)}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\]

Through a similar process, you can show that for \(0\leq y\leq \pi\)

\[\frac{d(\cos ^{-1}u)}{dx}=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\]

Inverse Tangent Function

If \(y=\tan ^{-1} u, then

\begin{align} u&=\tan y \\ \frac{du}{dx} &= \sec ^2y\frac{dy}{dx} \\ \frac{dy}{dx}&=\frac{1}{ \sec ^2y}\frac{du}{dx} \\ \frac{dy}{dx}&=\frac{1}{1+ \tan ^2y}\frac{du}{dx} \end{align}

Thus, the derivative is 

\[\frac{d(\tan ^{-1}u)}{dx}=\frac{1}{1+ u^2}\frac{du}{dx} \]


Example 1: Find the derivative of \[V=8\tan ^{-1}\sqrt{s}\]


\begin{align} V'&=8\frac{1}{1+(\sqrt{s})^2}\left(\frac{1}{2}s^{-\frac{1}{2}}\right) \\ &=4\frac{1}{(1+s)\sqrt{s}} \\&=\frac{4\sqrt{s}}{s(1+s)}\end{align}

Example 2: Find the derivative of \[p=\frac{3}{\cos ^{-1}2w}\]

Solution: We are going to derive the equation in the form \(p=3\left(\cos ^{-1}2w\right)^{-1}\) and use the chain rule

\begin{align} p'&=  -3\left(\cos ^{-1}2w\right)^{-2}\left(-\frac{1}{\sqrt{1-(2w)^2}}\right)(2) \\ &= \frac{6}{\left(\cos ^{-1}2w\right)^2\left(\sqrt{1-(2w)^2} \right)}\end{align}

Example 3: Find the second derivative of \[y=\tan ^{-1}2x\]

Solution: The first derivative is


Now find another derivative


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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
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