# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sqrt{\frac{1+x}{1-x}}$

Solution:

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

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

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

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

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

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

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

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

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

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

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$ and derivative of a constant is 0 .

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

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

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

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

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

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

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

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

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

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

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

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right)=\frac{1}{(1-\mathrm{x})^{\frac{3}{2}} \sqrt{1+\mathrm{x}}}$