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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function. Then the derivative of $f$ is denoted $f'$ and is defined by the mapping: $$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}.$$

Now suppose we also have the function $g:\mathbb{R}\rightarrow\mathbb{R}$.

Would it then be the case that "the derivative of $f$ with respect to $g$" (if this is a proper notion at all) is denoted $f'(g)$ and is defined by the mapping: $$f'\left(g(a)\right)=\lim_{x\rightarrow a}\frac{f\left(g(x)\right)-f(\left(g(a)\right)}{g(x)-g(a)}?$$

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Yes of course, given that $f(x)$ is differentiable in $b=g(a)$ and $g(x)$ is continuous at $x=a$ we have

$$\lim_{x\rightarrow a}\frac{f\left(g(x)\right)-f(\left(g(a)\right)}{g(x)-g(a)}=\lim_{y\rightarrow b}\frac{f(y)-f(b)}{y-b}=f'(b)=f'(g(a))$$

user
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  • @Subhajit above linked to another answer which states instead, "One should also note that we should define the derivative of $f$ with respect to $g$ as the following limit: $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{g(x+h)-g(x)}."$$ I think you and I are correct but do you know why that other person might have thought differently? –  Sep 22 '18 at 00:32
  • @dtcm840 The two are equivalent under the given hypotesis. – user Sep 22 '18 at 07:59