Find the derivative of $f$ at $2$

Question

Let $g$ be a real valued function defined on the interval $(-1,1)$ such that $$e^{-x}(g(x)-2e^x)=\int_0^x \sqrt{y^4+1}\,dy$$ for all $x\in (-1,1)$ and $f$ be another function such that $$f(g(x))=g(f(x))=x.$$ Then find the value of $$f'(2).$$

So, first I tried alot a then we tried it in an integration calculator but $\int_0^x \sqrt{y^4+1} \, dy$ is not possible. Next, since $f(g(x))=g(f(x))=x$ is given, therefore $f^{-1}=g$ and the vice versa. But I could not even find the dumbest idea how to do this question. Please help me out guys. Thanks!! Cheers!!

Answer 1

Just a small Hint: Derive $f(g(x))=x$ by chain rule: $$(x)'=1=(f(g(x)))'=f'(g(x))g'(x)$$ Hence $$f'(g(x))=\frac{1}{g'(x)}.$$ So you need to find an $x$ where $g(x)=2$ and calculate $g'(x)$.

Answer 2

Notice that evaluating the first equation in 0 yields $g(0)=2$.

You might then know that we are looking for $g'(0)$ because of this formula.

Then, by deriving the first expression using the fundamental theorem of calculus, we find: $$ -e^{-x}g(x)+g'(x)e^{-x}=\sqrt{x^4+1} $$ which, when evaluated in 0 yields: $$ g'(0)-g(0)=1 $$ Then, since $g(0)=2$, you may conclude easily that $f'(2)=\frac{1}{3}$.