The Inverse Function Theorem

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We see the theoretical underpinning of finding the derivative of an inverse function at a point.

There is one catch to all the explanations given above where we computed derivatives of inverse functions. To write something like $\ddx (e^y)=e^y\cdot y'$ we need to know that the function $y$ has a derivative. The Inverse Function Theorem guarantees this.

Inverse Function Theorem If $f$ is a differentiable function that is one-to-one near $a$ and $f'(a) \neq 0$ , then

(a): $f^{-1}(x)$ is defined for $x$ near $b=f(a)$ ,
(b): $f^{-1}(x)$ is differentiable near $b=f(a)$ ,
(c): last, but not least: $\eval {\ddx f^{-1}(x)}_{x=b} = \frac {1}{f'(a)}\qquad \text {where}\qquad b = f(a).$

We will only explain the last result. We know $f(f^{-1}(x)) = x,$ and now we use implicit differentiation (and the chain rule) to write $\begin{align*} \ddx f(f^{-1}(x)) &= f'(f^{-1}(x)) (f^{-1})'(x)\\ &=1. \end{align*}$ Solving for $(f^{-1})'(x)$ we see $(f^{-1})'(x) = \frac {1}{f'(f^{-1}(x))}.$ This is what we have written above.

It is worth giving one more piece of evidence for the formula above, this time based on increments in function, $\Delta f$ , and increments in variable, $\Delta x$ . Consider this plot of a function $f$ and its inverse:

Since the graph of the inverse of a function is the reflection of the graph of the function over the line $y=x$ , we see that the increments are “switched” when reflected. Hence we see that $\dfrac {\Delta f^{-1}}{\Delta {x}} = \dfrac {\Delta x}{\Delta f}=\dfrac {1}{\dfrac {\Delta f}{\Delta x}}.$ Taking the limit as $\Delta x$ goes to $0$ , we can obtain the expression for the derivative of $f^{-1}$ .
$\dfrac {d f^{-1}}{d{x}} = \dfrac {1}{\dfrac {d f}{d x}}$

The inverse function theorem gives us a recipe for computing the derivatives of inverses of functions at points.

Let $f$ be a differentiable function that has an inverse. In the table below we give several values for both $f$ and $f'$ : $\begin {array}{|c|c|c|}\hline x & f & f' \\ \hline \hline 2 & 0 & 2 \\ \hline 3 &1 &5 \\ \hline 4 & 3 & 0 \\ \hline \end {array}$ Compute $\ddx f^{-1}(x)\;\text {at $x=1$.}$

From the table above we see that $1 = f(\answer [given]{3}).$ Hence, by the inverse function theorem $\left (f^{-1}\right )' (1)= \frac {1}{f'(\answer [given]{3})} = \answer [given]{\frac {1}{5}}.$

If one example is good, two are better:

Let $f$ be a differentiable function that has an inverse. In the table below we give several values for both $f$ and $f'$ : $\begin {array}{|c|c|c|}\hline x & f & f' \\ \hline \hline 2 & 0 & 2 \\ \hline 3 & 1 &5 \\ \hline 4 & 3 & 0 \\ \hline \end {array}$ Compute $\left (f^{-1}\right )'(0)$

Note, $\left (f^{-1}\right )'(0) = \ddx f^{-1}(x)\;\text {at $x=0$.}$ From the table above we see that $0 = f(\answer [given]{2}).$ Hence, by the inverse function theorem $\left (f^{-1}\right )'(0) = \frac {1}{f'(\answer [given]{2})} = \answer [given]{\frac {1}{2}}.$

Finally, let’s see an example where the theorem does not apply.

Let $f$ be a differentiable function that has an inverse. In the table below we give several values for both $f$ and $f'$ : $\begin {array}{|c|c|c|}\hline x & f & f' \\ \hline \hline 2 & 0 & 2 \\ \hline 3 & 1 & 4 \\ \hline 4 & 3 & 0 \\ \hline \end {array}$ Compute $\eval {\ddx f^{-1}(x)}_{x=3}$

From the table above we see that $3 = f(\answer [given]{4}).$ Ah! But here, $f'(\answer [given]{4}) = \answer [given]{0}$ , so we have no guarantee that the inverse exists near the point $x=3$ , but even if it did the inverse would not be differentiable there.