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# Calculus

## 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 $$u$$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$

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 $$0\leq y\leq \pi$$

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

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 $\frac{d(\tan ^{-1}u)}{dx}=\frac{1}{1+ u^2}\frac{du}{dx}$ ## Examples Example 1: Find the derivative of $V=8\tan ^{-1}\sqrt{s}$ Solution: \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

$y'=\frac{2}{1+2x}$

Now find another derivative

$y''=-4\left(1+2x\right)^{-2}$