I came across this question:
The function $f(x) = e^x + x$ being differentiable and one to one, has a differentiable inverse $f^{-1}(x)$. Find the value of $\frac{d}{dx}(f^{-1})$ at the point $f(log(2))$.
I approached this as: $$ f^{-1}(f(x)) = x $$
So, $$ \frac{d}{dx}(f^{-1}(f(x))) = \frac{d}{dx}(x) $$
which simplifies to $$ \frac{d}{dx}(f^{-1}(f(x))) = 1 $$ Hence, $$ \frac{d}{dx}(f^{-1}(f(log(2)))) = 1 $$
But this does not seem right to me, because it does not matter if I find the derivated of $f^{-1}(f(x))$ at $f(log(2))$ or any other point, neither does it depend on the function, by above reasoning the answer should always be $1$. Where am I wrong?
You had it fine up till $$\frac{d}{dx}(f^{-1}(f(x))) = 1$$
but then you can't simply plug in $x = \log(2)$ since the value inside the derivative would then be a constant, for which the derivative is $0$. Instead, use the chain rule to get $$f'(x) \cdot (f^{-1})'(f(x)) = 1$$
Then dividing by $f'(x)$, this becomes $$(f^{-1})'(f(x)) = \frac{1}{f'(x)}$$
Now plug in $t = f(x) \to x = f^{-1}(x)$ to get $$(f^{-1})'(t) = \frac{1}{f'(f^{-1}(t))}$$
Edit: Since you are already given that the point at which to find the derivative is $f(\log(2))$, you know that $f^{-1}(t) = \log(2)$. Then the derivative at that point would be $$\frac{1}{e^{\log(2)}+1} = \frac{1}{3}$$ since $f'(x) = e^x + 1$