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: $$f'\left(g(a)\right)=\lim_{x\rightarrow a}\frac{f\left(g(x)\right)-f(\left(g(a)\right)}{g(x)-g(a)}?$$
Yes of course, given that $f(x)$ is differentiable in $b=g(a)$ and $g(x)$ is continuous at $x=a$ we have
$$\lim_{x\rightarrow a}\frac{f\left(g(x)\right)-f(\left(g(a)\right)}{g(x)-g(a)}=\lim_{y\rightarrow b}\frac{f(y)-f(b)}{y-b}=f'(b)=f'(g(a))$$