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

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

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

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