Given a function such as \(\tan x\), could you write \(\frac{d}{dx} \arctan x\) and \(\int \arctan x \; dx\), just from \(\tan x\), \(\frac{d}{dx} \tan x\) and \(\int \tan x \; dx\)? With some caveats, the inverse function theorem answers the former while the Legendre transformation answers the later.

I’ve writtten an updated / clearer version of this article in pdf. However, the web page is slightly more technical in presentation.

Derivatives of inverse functions and the inverse function theorem

Given a continuously differentiable function \(f: \mathbb{R} \to \mathbb{R}\) with \(f'(a) \neq 0\) at some point, the inverse function theorem states that there is some interval \(I\) with \(a \in I\) such that there exists a continuously differentiable inverse \(f^{-1}\) defined on \(f(I)\) such that

\[\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} \; \forall x \in I\]

The standard way to deduce the above formula would be to differentiate both sides of \(f^{-1}(f(x)) = x\). A simple example of this would be that the derivative of \(\ln x\) is \(1/e^{\ln x} = 1/x\).

Geometrically, given some function \(f(x) = y\) the inverse is just taking the plot of the function and reflecting it about the diagonal line \(y=x\). As such all tangent lines are also reflected along the diagonal line and hence the slope is inversed. Below we have the graph of \(e^x\) (the blue line) and the graph of \(\ln x\) (the red line). And you can see how the tangent line of \(e^x\) at \((2,e^2)\) is reflected along \(y=x\) to give the tangent line of \(\ln x\) at \((e^2, 2)\).

Another way of looking at it would be to subjugate \(S=\{(x, f(x)) \, \vert \, x \in I\}\), the plot of \(f(x)\), with a graph transformation \(\phi:(x,y) \to (\hat{x}, \hat{y}) = (y,x)\) and express \(\phi(S)\) as a plot \((\hat{x}, \hat{y}(\hat{x}))\). This way \(\hat{y}\) would be in effect the inverse of \(y\). We then wish to understand how \(\phi\) acts on \(\frac{dy}{dx}\) to give \(\frac{d\hat{y}}{d\hat{x}}\).

https://q.uiver.app/#q=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

We could then directly calculate

\[\begin{align*} \frac{d\hat{y}}{d\hat{x}}\vert_{\hat{x} = y(x)} &= \frac{d\hat{y}}{dx} / \frac{d\hat{x}}{dx} & \text{(Chain rule)} \\ &= 1 / \frac{dy}{dx} \end{align*}\]

The above is equivalent to taking the derivative of both sides of \(f^{-1}(f(x))= x\). However, this method is not only less ad hoc, but also generalises to other maps \(\phi: \mathbb{R}^2 \to \mathbb{R}^2\) (e.g. \(\phi:(r, \theta) \to (r \cos \theta, r \sin \theta)\)) . These ideas are explored much more deeply in Hydon’s book on Symmetry Methods for Differential Equations.

By calculating \(\frac{d}{dx} \tan x = \sec^2 x\), we now know that \(\frac{d}{dx} \arctan x = \frac{1}{\text{sec}^2(\arctan x)} = \frac{1}{1+x^2}\). So what about \(\int \arctan x \, dx\)? We shall do so without much rigour.

Integrals of inverse functions and the Legendre transformation

Geometrically, we could show that with \(f\) strictly monotone, we have

\[\int_{c}^d f^{-1}(y) \, dy + \int^{b}_a f(x) \, dx = bd-ac\]

This is a result by Laisant in 1905. Here’s a proof-without-words.

By Jonathan Steinbuch - Own work, CC BY-SA 3.0, Link

To find an explicit formula for \(\int f^{-1}(y)\), we could do the following. Suppose we’re interested in finding \(\int \arctan x \, dx\). Notationally, it is the easiest to let \(y = \int \tan x \, dx\) up to some constant such that \(\frac{dy}{dx} = \tan x\) is the function we want to invert. Let \(S\) be the plot of \(\frac{dy}{dx}\), i.e. \(\{(x, \frac{dy}{dx}) \vert \, x \in I\}\) for some appropiate interval \(I\). Applying the map \(\phi: (x,y) \to (\hat{x}, \hat{y}) = (y,x)\), we assume the existence of some \(\hat{y}\) such that its derivative is the plot of \(\phi(S)\). Now \(\hat{y} = \int \arctan \hat{x} \, d\hat{x}\) which is what we seek.

https://q.uiver.app/#q=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

For the sake of clarity we write \(\frac{dy}{dx}\vert_{\hat{x} = u}\) as \(y'(u)\) and \(\frac{d\hat{y}}{d\hat{x}}\vert_{\hat{x} = u}\) as \(\hat{y}'(u)\). So

\[\begin{align*} \hat{y} &= \int \hat{y}'(\hat{x}) d\hat{x} \\ &= \hat{x} \times \hat{y}' - \int \hat{x} \times \hat{y}'(\hat{x}) d\hat{x} & (\text{by parts}) \\ &= \hat{x} \times \hat{y}' - \int \hat{x}(x) \times \hat{y}'(\hat{x}(x)) \times \hat{x}'(x) dx & (\text{sub }\hat{x} = y'(x))\\ &= \hat{x} \times \hat{y}' - \int \hat{x}(x) \times \big[\frac{d}{dx}\hat{y}'\big] dx \\ &= \hat{x} \times \hat{y}' - \int \hat{x}(x) dx & (\hat{y}'(\hat{x}(x)) = x) \\ &= \hat{x} \times \hat{y}' - \int y'(x) dx & (\text{by def of } \hat{x})\\ &= \hat{x} \times \hat{y}'(\hat{x}) - y(x) + C \\ &= \hat{x} \times \hat{y}'(\hat{x}) - y(\hat{y}'(\hat{x})) + C & (\text{by def of } \hat{x}) \end{align*}\]

As an example, if we let \(y(x) = e^x\) then \(y'(x) = e^x\), \(\hat{y}'(\hat{x}) = \ln (\hat{x})\) and so repeating what we’ve done above,

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We have

\[\begin{align*} \hat{y} &= \int \ln \hat{x} \, d\hat{x} \\ &= \hat{x} \times \ln(\hat{x}) - \int \hat{x} \times (\ln \hat{x}) d\hat{x}\\ &= \hat{x} \times \ln(\hat{x}) - \int e^x \times \ln(e^x) \times e^x dx \\ &= \hat{x} \times \ln(\hat{x}) - \int e^x \times \frac{d}{dx} (\ln(e^x)) dx \\ &= \hat{x} \times \ln(\hat{x}) - \int e^x dx \\ &= \hat{x} \times \ln(\hat{x}) - e^x + C \\ &= \hat{x} \times \ln(\hat{x}) - e^{\ln (\hat{x})} + C \\ &= \hat{x} \times \ln(\hat{x}) - \hat{x} + C \\ \end{align*}\]

Finally, just by using the formula \(\hat{y} = \hat{x} \times \hat{y}'(\hat{x}) - y(\hat{y}'(\hat{x})) + C\), if we let \(y = - \ln \vert \cos(x) \vert\), then we have \(y' = \tan(x)\) and \(\hat{y}' = \arctan(x)\) so

\[\begin{align*} \int \arctan \hat{x} &= \hat{x} \arctan \hat{x} - \ln \vert \cos(\arctan(\hat{x}))\vert + C\\ &= \hat{x} \arctan \hat{x} - \frac{1}{2} \ln \big( \frac{1}{1+\hat{x}^2}\big) + C \end{align*}\]

as expected.

The map \(y(x) \to \hat{y}(\hat{x}) := \hat{x} \hat{y}'(\hat{x}) - y(\hat{y}'(\hat{x}))\) is called the Legendre transformation, which has wide applications in areas of physics. But mathematically, one could think of it as an analogue to the inverse function theorem: It tells you how the inverse map acts on integrals. Much of this section is inspired from a wikipedia article on integrals of inverse functions.

Summary

https://q.uiver.app/#q=WzAsNixbMCwxLCJmKHgpIl0sWzAsMiwiZicoeCkiXSxbMCwwLCJGKHgpIl0sWzMsMSwiZyh5KTo9Zl57LTF9KHkpIl0sWzMsMCwiRyh5KSJdLFszLDIsImcnKHkpIl0sWzAsMSwiXFxmcmFje2R9e2R4fSJdLFsyLDAsIlxcZnJhY3tkfXtkeH0iXSxbMCwzXSxbNCwzLCJcXGZyYWN7ZH17ZHl9Il0sWzMsNSwiXFxmcmFje2R9e2R5fSJdLFsxLDUsIlxcdGV4dHtJbnZlcnNlIEZ1bmMuIFRobS59Il0sWzIsNCwiXFx0ZXh0e0xlZ2VuZHJlIFRyYW5zZm9ybX0iXV0=

with the following relations

\[\begin{align*} \text{IFT:} \quad & g'(y) = \frac{1}{f'(g(y))} \\ \text{Legendre:} \quad & G(y) = y\times g(y) - F(g(y)) + C \end{align*}\]