Question

If $y={\cos }^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$ then $\frac{dy}{dx}=$

Answer 1

Consider the given equation.

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

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

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

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

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

$\frac{dy}{dx}=-\frac{(x^2+1)}{\sqrt{x^4+1+2x^2-(x^4+1-2x^2)}}$

$\frac{dy}{dx}=-\frac{(x^2+1)}{\sqrt{4x^2}}$

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

Hence, this is the answer.