Q.If $y = f(x)$ is a differentiable function of $x$ such that inverse function $x = f^{-1}(y)$ exists, then prove that $x$ is a differentiable function of $y$ and $\dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}$ where $\dfrac{dy}{dx} \ne 0$. Hence find $\dfrac{d}{dx}\left(\tan^{-1}x\right)$.

If $y=f(x)$ is a one-one, onto (hence invertible) differentiable function with $\dfrac{dy}{dx}\ne0$, then its inverse $x=f^{-1}(y)$ is also differentiable, and its derivative is the reciprocal of the original derivative: $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}$, evaluated at the corresponding point. Equivalently, if $g=f^{-1}$, then $g'(y)=\dfrac{1}{f'(x)}$ where $y=f(x)$. This can be proved two ways: (1) directly from increments, since $\dfrac{\delta x}{\delta y}\cdot\dfrac{\delta y}{\delta x}=1$, so taking the limit gives $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}$; or (2) by differentiating the identity $f^{-1}[f(x)]=x$ using the chain rule, which gives $(f^{-1})'[f(x)]\cdot f'(x)=1$. To use this in practice: write $y=f(x)$, invert to get $x=f^{-1}(y)$ explicitly in terms of $y$, and differentiate $x$ with respect to $y$ directly — or find $dy/dx$ first and take its reciprocal. This idea is what lets us derive the standard derivatives of inverse trigonometric functions and any other invertible function without a fresh first-principles proof each time. …