Differentiate the following functions with respect to x :
Question:

Differentiate the following functions with respect to $x$ :

$\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$

Solution:

Let $y=\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{a^{2}-x^{2}}{a^{2}+x^{2}}\right)^{-\frac{1}{2}}\left[\frac{\left(a^{2}+x^{2}\right)(0-2 x)-\left(a^{2}-x^{2}\right)(0+2 x)}{\left(a^{2}+x^{2}\right)^{2}}\right]$

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \mathrm{xa}^{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{-\frac{1}{2}}}{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)^{\frac{3}{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-2 x a^{2}}{\left(a^{2}+x^{2}\right)^{\frac{3}{2}}\left(a^{2}-x^{2}\right)^{\frac{1}{2}}}$

$\therefore \frac{d y}{d x}=\frac{-2 x a^{2}}{\left(a^{2}+x^{2}\right)^{\frac{3}{2}} \sqrt{a^{2}-x^{2}}}$

Thus, $\frac{d}{d x}\left(\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}\right)=\frac{-2 x a^{2}}{\left(a^{2}+x^{2}\right)^{\frac{3}{2}} \sqrt{a^{2}-x^{2}}}$