Differentiation of Inverse Functions

Differentiating inverse functions is quite simple. To do this, you only need to learn one simple formula shown below:

$$\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(y)},y=f^{-1}(x)$$

That was quite simple, wasn't it? However, when the problem is a little tricky, it might get confusing to decide which variable should be substituted into $x$ and which into $y.$ This is why an understanding of the proof is essential. When it comes to inverse functions, we usually change the positions of $y$ and $x$ in the equation. Of course, this is because if $y=f^{-1}(x)$ is true, then $x=f(y)$ is also true. The proof for the formula above also sticks to this rule.

Prove that the derivative of $y=f^{-1}(x)$ with respect to $x$ is $\frac{1}{f'(y)}.$

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We know that $x=f(y).$ Differentiating both sides with respect to $y$ gives

$$\begin{aligned} \frac{dx}{dy}&=\frac{d}{dy}f(y)\\ \frac{dx}{dy}&=f'(y)\\ \Rightarrow \frac{dy}{dx}&=\frac{1}{f'(y)}.\ _\square \end{aligned}$$

Don't forget that $x$ and $y$ are respectively the input and output of the function $y=f^{-1}(x)$ in this formula. A graphical demonstration will give us more insight into this:

From the figure above, we can see that the points $(x,y)$ and $(y,x)$ are symmetrical about the line $y=x$ and so are the tangents of those points. Thus we can conclude from the figure that $\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(y)}.$

If $f(x)=x^2,$ what is the derivative of $f^{-1}(x)$ at $x=4?$

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Since $f(2)=4,$ we know that $f^{-1}(4)=2.$ Hence we have

$$\frac{d}{dx}f^{-1}(4)=\frac{1}{f'(2)}= \left. \frac{1}{2x} \right | _{x=2}=\frac{1}{4}.\ _\square$$

If $f(x)=\ln x,$ what is the derivative of $f^{-1}(x)$ at $x=2?$

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Since $f(e^2)=2,$ we know that $f^{-1}(2)=e^2.$ Hence we have

$$\frac{d}{dx}f^{-1}(2)=\frac{1}{f'(e^2)}= \left. \frac{1}{\frac{1}{x}} \right |_{x=e^2}=e^2.\ _\square$$

If $f(x)=\sin x\ \ \left(-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}\right),$ what is the equation of the tangent of $y=f^{-1}(x)$ at $x=\frac{1}{2}?$

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Since $f\left(\frac{\pi}{6}\right)=\frac{1}{2},$ we know that $f^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}.$ Hence we have

$$\frac{d}{dx}f^{-1}\left(\frac{1}{2}\right)=\frac{1}{f'\left(\frac{\pi}{6}\right)}=\left. \frac{1}{\cos x}\right |_{x=\frac{\pi}{6}}=\frac{2}{\sqrt{3}}.$$

This implies that the slope of the tangent is $\frac{2}{\sqrt{3}}.$ We know that the tangent passes through the point $\left(\frac{1}{2},\frac{\pi}{6}\right).$ Therefore, the equation of the tangent is

$$\begin{aligned} y&=\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)+\frac{\pi}{6}\\ y&=\frac{2}{\sqrt{3}}x-\frac{\sqrt{3}}{3}+\frac{\pi}{6}.\ _\square \end{aligned}$$