Q.Find the derivative of the function $y=f(x)$ using the derivative of the inverse function $x=f^{-1}(y)$: $y=2-\sqrt{x}$

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.

"Derivative of inverse function formula dx/dy" and "continuity and differentiability important questions class 12" recur around the NCERT/CBSE Class 12 Mathematics curriculum, since this result underlies the standard derivatives of every inverse trigonometric function tested in board exams and JEE Main. Being comfortable with both proof routes shown here, direct increments and the chain-rule identity, gives students flexibility when a competitive-exam question asks for a derivation rather than just the formula.