Derivative of $f \left( g^{-1} (x) \right)$ using only the first principle

Question

I am trying to find the derivative of $f \left( g^{-1} (x) \right)$. I know that it is given by $$f \left( g^{-1} (x) \right)'=\frac{f' \left( g^{-1} (x) \right)}{ g' \left( g^{-1} (x) \right)}$$ using chain rule.

But I am not able to prove it using first principle.

How can I do that using the first principle ?

Answer 1

Enforcing the change of variables $\color{black}{y=g^{-1}(x)~~~and ~~~~b=g^{-1}(a)} $ then $ y\to b$ as $x\to a$ and hence by firs principle we have

$$\begin{align} \left(f(g^{-1})\right)'(a) &=\lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{x-a}\\&= \lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{\color{blue}{g^{-1}(x)-g^{-1}(a)}}\cdot\frac{\color{blue}{g^{-1}(x)-g^{-1}(a)}}{x-a}\\&=\lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{\color{blue}{g^{-1}(x)-g^{-1}(a)}}\cdot\frac{\color{blue}{g^{-1}(x)-g^{-1}(a)}}{\color{red}{g(g^{-1})(x)-g(g^{-1})(a)}} \\&=\lim_{y\to b}\frac{f(y)-f(b)}{\color{blue}{y-b}}\cdot\frac{\color{blue}{y-b}}{\color{red}{g(y)-g(b)}} = \color{black}{f'(b)\cdot \frac{1}{g'(b)}}\\&= \color{brown}{ \frac{f'(g^{-1}(a))}{g'(g^{-1}(a))}}\end{align}$$

Answer 2

Suppose that $g$ is an invertible function. Then $g^{-1}\circ g = g\circ g^{-1}=id$.

Notice that $g(g^{-1}(x))=x$ and differentiate both sides using the chain rule on the left: $$g'(g^{-1}(x)).(g^{-1}(x))'=1 $$ Thus $$(g^{-1}(x))'=\frac{1}{g'(g^{-1}(x))} $$

Now differentiate $f(g^{-1}(x))$ using the chain rule again to get $$(f(g^{-1}(x)))'=f'(g^{-1}(x))(g^{-1}(x))' $$

But we've calculated the second term above so $$ (f(g^{-1}(x)))'=\frac{f'(g^{-1}(x))}{g'(g^{-1}(x))} $$