I am looking for verification on proving that for $$ f(x) = \int_1^x \frac{dt}{t} $$ the derivative of the inverse function is the inverse function itself: $$ f^{-1}(y) = \frac{d}{dy}f^{-1}(y). $$
Obviously the statement is true because
- $f(x) = \log(x)$ and therefore $f^{-1}(y) = \exp(y)$
- And because $\exp(y) = \frac{d}{dy}\exp(y)$.
However, I want to prove the fact using only the definitions of $f$ and $f^{-1}$.
So far I know that by the fundamental theorem of calculus $$ f'(x) = 1/x. $$
And by the inverse function theorem I also know that $$ \frac{d}{dy} f^{-1}(y) = \frac{1}{f'(x)}~ \text{ such that } f(x)=y. $$
And therefore $$ \frac{d}{dy} f^{-1}(y) = \frac{1}{f'(f^{-1}(y))} = f^{-1}(y). $$
So I guess that solves it... ?
Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,\infty)$ so the inverse function theorem actually applies.