What does rate of the change of a function with respect to another function at a point mean?

Question

My question is inspired from this problem:

What is rate of the change of $\sqrt{x^2+16}$ with respect to changes of $\dfrac x{x-1}$ at the point $x=3$?

It appears that we should evaluate the following expression for $x=3$:$$\dfrac{d\left(\sqrt{x^2+16}\right)}{d\left(\dfrac{x}{1-x}\right)}=\dfrac{\dfrac d{dx}\left(\sqrt{x^2+16}\right)}{\dfrac d{dx}\left(\dfrac{x}{1-x}\right)}$$

Which leads to $-2.4$.

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It was an example and my question is not about this specific problem. I'm wondering what does rate of change of a function with respect to another function actually means. I know that rate of change of $f(x)$ with respect to its variable is $\dfrac d{dx}(f(x))$ and at the point $x=x_0$, it represents the slope of the line tangent to it, but I don't have any intuition about rate of the change with respect to another function at a given point.

Answer 1

It basically describes how much $f(x)$ changes per unit change in $g(x)$. When $g(x)=x$ it becomes far more intuitive, but it's the same concept.

You can see $g(x)$ as an independent variable; this way, $df/dg$ at a point $g_0$ represents the slope of the tangent line to the graph of the function $g\mapsto f(g^{-1}(x))$ at the point $g=g_0$ (provided that $x\mapsto g(x)$ is bijective, otherwise $g^{-1}$ denotes any of the branches of its inverse relation).

Answer 2

The rate of change of a function $f(x)$ with respect to another function $g(x)$ at the point where $x=a$ simply means the ratio of the change in $f(x)$ caused by a small change$\big{(}$tending to $0$$\big{)}$ in $g(x)$ around $g(x)=g(a)$.

You can think of it as the slope of the tangent line to the graph plotted between $g(x)$ on the $X$-axis and $f(x)$ on the $Y$-axis at the point $\left(g(a),f(a)\right)$. Note that when $g(x)=x$, this rate of change simply becomes the derivative/rate of change or $f(x)$ w.r.t. $x$ at $\left(a,f(a)\right)$, which is what one is first introduced to when learning about derivatives.

P.S. Let me also prove why we divide by the derivatives of $f(x)$ and $g(x)$ at $x=a$(this is only applicable when $g'(a)\ne0$, see below) to find this rate of change$\bigg{(}$denoted by $\frac{\mathrm df}{\mathrm dg}$$\bigg{)}$. It can also be thought of as an application of the chain rule.

$$\begin{align}\frac{\mathrm df}{\mathrm dg}&=\lim_{g(x)\to g(a)}\frac{f(x)-f(a)}{g(x)-g(a)}\\&=\lim_{x\to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}\\&=\frac{\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}(\text{if }\lim_{x\to a}\frac{g(x)-g(a)}{x-a}=g'(a)\ne0)\\&=\frac{f'(a)}{g'(a)}\end{align}$$

If $g'(a)=0$, two cases arise:

$\textbf{Case 1, $f'(a)=0$:}$ In this case, $\frac{\mathrm df}{\mathrm dg}$ is given by:

$$\frac{\mathrm df}{\mathrm dg}=\lim_{x\to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}$$

which is the same as

$$\frac{\mathrm df}{\mathrm dg}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}$$

Example, the rate of change of $\cos x$ w.r.t $x^2$ at $x=0$ is $\frac12$.

Note that the above limit may or may not exist. Example, the rate of change of $\cos x$ w.r.t $x^3$ at $x=0$ is not defined.

$\textbf{Case 2, $f'(a)\ne0$:}$ In this case, the derivative does not exist since one of the limits will be $+\infty$ and the other will be $-\infty$ as $\lim_{x\to0^+}\frac1{x}=+\infty$ and $\lim_{x\to0^-}\frac1{x}=-\infty$.

Example, the rate of change of $e^x$ w.r.t $\cos x$ at $x=0$ is not defined.

Answer 3

Never heard of that notation but you can interpreted as follows. Since $$ \frac{\partial f}{\partial x}(x)=\lim_{t\to 0}\frac{f(x+t)-f(x)}{t} = (f\circ \lambda)'(0)$$ where $\lambda$ is the parametrization $\lambda(t)=x+t$. The rate of change of $f$ with respect of $g$ could interpreted as taking $g(t)$ instead of $\lambda(t)$ as the parametrization. That's my best interpretation.

Answer 4

The rate of change of $f$ with respect to $g$ is, just like the rate of change of $f$ with respect to $x$, the limit as $\Delta g(x)$ goes to zero of $\frac{\Delta f(x)}{\Delta g(x) }$. This isn't well-defined for all $g$. I'm not sure the exact necessary and sufficient conditions, but it's probably something like "it's well defined at $x_0$ if there is neighborhood around $x_0$ such that $g'$ is continuous and nonzero". Then in the limit, $\Delta g$ = $g'\Delta x$, so $\frac{\Delta f(x)}{\Delta g(x) } = \frac{\Delta f(x)}{g'\Delta x }= \frac{1}{g'} \frac{\Delta f(x)}{\Delta x }=\frac{1}{g'}{f'}=\frac {f'}{g'}$ .