$$ \newcommand{\d}[0]{ \text{d} } $$

The gist

The inverse function theorem is a an elegant observation stemming from the symmetry of functions and their inverse along the $y=x$ axis.

Consider a function $y = f(x)$ and its inverse $x = g(y) $, the inverse theorem states that you can compute $ g'(y) $ purely based on $f'$ without prior knowledge of $g$ (notice, but temporarly ignore that the equality below links two seemingly unrelated variables):

$$ g'({\color{red} y }) = \frac{1}{f'({\color{green} x })} $$

This is a pretty neat fact, it means when you can't find the inverse $g$ you can still easily deduce its tangent $g'$ from the original function $f$.

Geometric interpretation

The crux is that the inverse of the tangent $ g' $ is the mirrored version of $f' $, which allows us to say $ g' = 1 / f' $. So in practice, let's set x to some value and compute the tangent $f'$ at some point $(x, \ y = f(x))$. The the inverse theorem then tells us, $g' = 1/f' $, however, where do you think this tangent should lie $(\cdots, \cdots)$ ? Certainly not at $(x, \ y)$! Well, it actually goes through the mirrored point $ (y, \ x) $. Just switching coordinates gives you the location of the inverse tangent $g'$. What's more is that just setting $x$ allows us to know $y = f(x)$, nowhere did we need $g$. This can be summarized by the equality below:

$$ g'(f(x)) = \frac{1}{f'(x)} $$

Let's rephrase the above to help things sink in. We want to find the tangent of $f$ and $g$ but only given a variable $x$. So far we know that $g'({\color{red} y }) = {1} / {f'({\color{green} x })}$ but how do we deduce the right ${\color{red} y }$ that corresponds to the tangent $g'$ of the inverse $g$? Well this is given by $ y = f(x) $.

Deriving the inverse theorem

Using the chain rule and the fact that $g(f(x)) = x$ it's simple to derive:

$$ \begin{aligned} g\big ( { \color{purple} f(x)} \big ) & = x \\ \Rightarrow \Big [ g\big ({ \color{purple} f(x)}\big) \Big ]' & = \Big [ x \Big ]'\\ g'\big ({ \color{purple} f(x)}\big) . f'(x) & = 1 \\ g'\big ({ \color{purple} f(x)}\big) & = \frac{1}{f'(x)} \\ \end{aligned} $$

Variation

By definition of the inverse $f$ and $g$ can be swapped:

$$ \begin{aligned} f\big (g(x) \big ) & = x \\ \Rightarrow f'\big ({ g(y)}\big) &= \frac{1}{g'(y)} \end{aligned} $$

Leibniz tricks

This one is neat, when using Leibniz notation to designate derivatives, it is very intuitive to get back the inverse theorem:

$$ \begin{gathered} \frac{ \d x}{\d {\color{red} y}} = \frac{1}{{\d y}\over{\d {\color{green} x}}} \Annotation{ \text{ thinking in terms rate change between variables: } \\ x = g(y) \text{ and } y = f(y)} \end{gathered} $$

We can see the above trick as renaming $f:y$ and $g:x$ :

$$ \begin{aligned} \frac{ \d g}{\d y} \quad & = \quad \frac{1}{{\d f}\over{ \d x}}\\ \frac{ \d \big( g(y) \big) }{\d y} & = \frac{1}{{\d\big(f(x)\big)}\over{\d x}} \\ g'(y) \quad & = \quad \frac{1}{ f'(x) } \\ \end{aligned} $$

Notice also the notation is a tad abusive because the equality only holds because we have hidden information:

\[ \frac{ \d x}{\d {\color{red} y}} = \frac{ \d x(\text{...}) }{\d {\color{red} y}} = \frac{ \d x( y ) }{\d {\color{red} y}} = \frac{ \d x( {\color{red} y} = f({\color{green} x}) ) }{\d {\color{red} y}} = g'( f({\color{green} x})) = \frac{1}{{\d y}\over{\d {\color{green} x}}} = \frac{1}{ f'({\color{green} x}) } \]

Consider $h_i: \mathbb R \rightarrow \mathbb R$; when making use of the Leibniz trick, keep in mind this change the input variable you use to evaluate the expression:

$$ h_1({\color{red} y}) = \frac{ \d x}{\d {\color{red} y}} $$ $$ h_2({\color{green} x}) = \frac{1}{{\d y}\over{\d {\color{green} x}}} $$

But that might turn out ok since we can always find a $y$ according to an $x$ with the original function $y = f(x)$:

$$ h_1\big (f({\color{green} x}) \big) = h_2({\color{green} x}) = \frac{1}{{\d y}\over{\d {\color{green} x}}} $$ $$ h_2\big (g({\color{red} y}) \big) = h_1({\color{red} y}) = \frac{ \d x}{\d {\color{red} y}} $$

Proper theorem

This is the exact definition of the inverse theorem.

If $f$ is differentiable at $c$, $f'$ is continuous at $c$, and $f'(c) \neq 0$, there is an interval $I$ containing $c$ on which $f^{-1}$ exists and is differentiable. Moreover:

$$ {f^{\scriptscriptstyle -1}}^{\displaystyle'}\big (f(c) \big) = \frac{1} {f^{\displaystyle \prime} (c)} $$

Example

Consider:

$$ f(x) = e^x + x^3 + x $$

The inverse $f^{-1}$ does not have a nice closed form expression but we can still compute $f'^{-1}$ if we set $c$; for instance $c= 0$

- $f(0) = 1$
- $f'(x) = e^x + 3 x^2 + 1$
- $f'(0) = 1 + 1 = 2$

$$ {f^{\scriptscriptstyle -1}}^{\displaystyle'}\big(f(0)\big) = {f^{\scriptscriptstyle -1}}^{\displaystyle'}(1) = \frac{1} {f^{\displaystyle \prime} (0)} = \frac{1} {2} $$

The crux really lies in the fact we evaluate ${f^{\scriptscriptstyle -1}}^{\displaystyle'}(1)$ at 1 and not at 0.

Reference

• Inverse theorem video explanation (University of Toronto Mississauga)
• Khan academy explanation

Images

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