Question

If $y=\sqrt{\frac{1-x}{1+x}}$, then $\frac{dy}{dx}$ equals -

Answer 1

We have,

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

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\times \left(\frac{(1+x)(0-1)-(1-x)(0+1)}{{(1+x)}^2}\right)$

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

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

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

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

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

Hence, this is the answer.