# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

Solution:

Let $y=\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

On differentiating $y$ with respect to $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\right]$

We have $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-\frac{x^{2}}{x^{2}+a^{2}}}} \frac{d}{d x}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{\frac{x^{2}+a^{2}-x^{2}}{x^{2}+a^{2}}}} \frac{d}{d x}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{\frac{a^{2}}{x^{2}+a^{2}}}} \frac{d}{d x}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)$

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}}\left[\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}\right)}{\left(\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}\right)^{2}}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}}\left[\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\right]^{\frac{1}{2}}}{\mathrm{x}^{2}+\mathrm{a}^{2}}\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\sqrt{x^{2}+a^{2}} \times 1-x\left(\frac{1}{2}\left(x^{2}+a^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(x^{2}+a^{2}\right)\right)}{x^{2}+a^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\sqrt{x^{2}+a^{2}}-x\left(\frac{1}{2}\left(x^{2}+a^{2}\right)^{\frac{1}{2}} \frac{d}{d x}\left(x^{2}+a^{2}\right)\right)}{x^{2}+a^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\sqrt{x^{2}+a^{2}}-\frac{x}{2 \sqrt{x^{2}+a^{2}}}\left(\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}\left(a^{2}\right)\right)}{x^{2}+a^{2}}\right]$

We have $\frac{d}{d x}\left(x^{2}\right)=2 x$ and derivative of a constant is 0 .

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}}\left[\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}-\frac{\mathrm{x}}{2 \sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}(2 \mathrm{x}+0)}{\mathrm{x}^{2}+\mathrm{a}^{2}}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}{\mathrm{a}}\left[\frac{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}-\frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}}{\mathrm{x}^{2}+\mathrm{a}^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{\frac{\left(\sqrt{x^{2}+a^{2}}\right)^{2}-x^{2}}{\sqrt{x^{2}+a^{2}}}}{x^{2}+a^{2}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{\sqrt{x^{2}+a^{2}}}{a}\left[\frac{x^{2}+a^{2}-x^{2}}{\sqrt{x^{2}+a^{2}}\left(x^{2}+a^{2}\right)}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{a}}\left[\frac{\mathrm{a}^{2}}{\mathrm{x}^{2}+\mathrm{a}^{2}}\right]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{\mathrm{x}^{2}+\mathrm{a}^{2}}$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left[\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right)\right]=\frac{\mathrm{a}}{\mathrm{x}^{2}+\mathrm{a}^{2}}$