Let $y=f(x)$, and suppose that $f$ has inverse function $g$, so that $x=g(y).$ If $f$ and $g$ are differentiable, then

Note that the derivatives here are evaluated at different points.

Geometrical interpretation. The tangent to the graph of $f$ at $(x,y)$ has gradient $f^{\prime}(x)$; so the reflection of this tangent in the line $y=x$ has gradient $1/f^{\prime}(x)$. But the graph of $g$ is the reflection of the graph of $f$ in the line $y=x$, so the tangent to the graph of $g$ at $(y,x)$ has gradient $1/f^{\prime}(x)$.

At the end of a calculation for the derivative of an inverse function, we need to express the answer as a function of $x$ by substituting $y=f(x).$ The inverse function rule is really a special case of the chain rule.

Let $y=f(x)$, so $x=g(y)=g(f(x))$, and differentiate to get

so $f^{\prime}(x)\neq 0$ and