Skip to Main Content
It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.


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


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