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 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 $f^{-1}$.

\[ \dd [f^{-1}]{x} = \dfrac {1}{ \dd [f]{x} } \]

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, three 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

\[ \left (f^{-1}\right )'(0) = \frac {1}{f'(\answer [given]{2})} = \answer [given]{\frac {1}{2}}. \]

The function $g(x) = x^2 + 3$ is one-to-one when restricted to the interval $(0, \infty )$. Let $g^{-1}(x)$ be the inverse of this restriction. Find an equation of the line tangent to the graph of $y=g^{-1}(x)$ at the point $(7, 2)$.

Since $g(2) = (2)^2 + 3 = 7$ we know that $(7, 2)$ is referring to the point on the graph of $g^{-1}$. In order to find the slope of the line, we must calculate $\left ( g^{-1}\right )'(7)$.

By the inverse function theorem we know:

\begin{align*} \left ( g^{-1}\right )'(7) &= \dfrac {1}{g'\left ( g^{-1}(7) \right )}\\ &= \dfrac {1}{g'\left ( 2 \right )}. \end{align*}

Since $g'(x) = 2x$, this gives $\left ( g^{-1}\right )'(7) = \dfrac {1}{4}$.

An equation of the tangent line can be given as:

\[ y-2 = \dfrac {1}{4}(x-7) \]

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.