Differentiation of one function with respect to another function

The chain rule (Theorem 10.5) differentiates a composite function with respect to the independent variable $x$. A closely related — and equally useful — question is: given two functions $f(x)$ and $g(x)$ of the same variable $x$, how does $f$ change as $g$ changes, i.e. what is $\dfrac{df}{dg}$?

If $f$ and $g$ are both differentiable functions of $x$ and $g'(x)=\dfrac{dg}{dx}\ne0$, then

This is proved exactly like the parametric-derivative formula of §10.4.6, treating $x$ itself as the connecting "parameter" between $f$ and $g$: $\dfrac{df}{dg}=\dfrac{df/dx}{dg/dx}$ is just the chain rule read the other way around, dividing one derivative-with-respect-to-$x$ by another.

Worked illustration. Find the derivative of $x^x$ with respect to $x\log x$. Let $u=x^x$ and $v=x\log x$. From §10.4.4, $\log u = x\log x$, so (by the same logarithmic-differentiation step) $\dfrac{du}{dx} = u(1+\log x) = x^x(1+\log x)$; and directly, $\dfrac{dv}{dx}=1\cdot\log x+x\cdot\dfrac1x = 1+\log x$ (product rule). Then

— the $(1+\log x)$ factors cancel exactly, leaving a strikingly clean result.

Worked illustration — inverse-trig pair. Find the derivative of $\tan^{-1}\!\left(1+x^2\right)$ with respect to $\sqrt{x^2+1}$. Let $f(x)=\tan^{-1}(1+x^2)$ and $g(x)=\sqrt{x^2+1}$. By the chain rule, $f'(x) = \dfrac{2x}{1+(1+x^2)^2}$ and $g'(x) = \dfrac{x}{\sqrt{x^2+1}}$. Then …