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


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 $$ (f(g^{-1}(x)))'=\frac{f'(g^{-1}(x))}{g'(g^{-1}(x))} $$
