Q.Differentiate $\cos^{-1}\dfrac{1-x^2}{1+x^2}$ w.r.t. $\tan^{-1}x$.

Given two functions $u=f(x)$ and $v=g(x)$, both differentiable functions of $x$, the derivative of $u$ with respect to $v$ (rather than with respect to $x$) is defined as $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}$, provided $dv/dx\ne0$. This is really the parametric-derivative idea applied with $x$ itself playing the role of the parameter: think of $u$ and $v$ as both depending on the 'parameter' $x$. To compute it, differentiate $u$ and $v$ separately with respect to $x$ in the usual way, then divide the two derivatives. This technique is especially useful when $u$ and $v$ are both composite/inverse-trig expressions that individually look complicated, because after differentiating each with respect to $x$, the ratio …