Differentiate the following functions with respect to x :

Solution:

Let $y=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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