Differentiate the following functions with respect to x :

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

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

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

We know $\frac{d}{d x}(\sin x)=\cos x$

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{4 x}{\left(1-x^{2}\right)^{2}}\right]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4 \mathrm{x}}{\left(1-\mathrm{x}^{2}\right)^{2}} \cos \left(\frac{1+\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right)$

Thus, $\frac{d}{d x}\left[\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]=\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$