Inverse function and fundamental theorem of calculus

Question

Let $f$ be the function defined by $$f(x) = \int_1^x \frac{1+|\sin xt^2|}{2}dt$$ which has an inverse function denoted by $g$. Find the derivative $g'(0)$

When $f(x) = 0$, I see that $x = 1$ is a solution. So I need to find $g'(0)=\frac{1}{f'(1)}$.

By the fundamental theorem of calculus, $$f'(x) = \frac{1+|\sin x^3|}{2}$$ $$f'(1)=\frac{1+\sin 1}{2}$$

However there is no such answer (the question is in the form of MCQ). I wonder where did I go wrong. Thanks.

Answer 1

$\textbf{Hint}$: Use the substitution $\sqrt{x}t = s$

$$f(x) = \frac{1}{2\sqrt{x}}\int_{\sqrt{x}}^{\sqrt{x^3}}1+|\sin(s^2)|\:ds$$

and now apply chain rule.

Answer 2

Let $f(x)=\frac12\int_1^x \left(1+|\sin(xt^2)|\right)\,dt$. Then, the derivative of $f$ at $x=1$, $f'(1)$, is given by

$$\begin{align} f'(1)&=\lim_{h\to 0}\frac{f(1+h)-\overbrace{f(1)}^{=0}}{h}\\\\ &\lim_{h\to 0}\frac{\frac12\int_1^{1+h}\left(1+|\sin((1+h)t^2)|\right)\,dt}{h}\\\\ &=\frac{1+\sin(1)}{2} \end{align}$$

And we are done!