How to find $g(x)$ here? If $$f(x)=\int_0^x \frac{dt}{\sqrt{1+t^4}} $$
I have been asked to find $g'(0)$. where $g=f^{-1}$
to do this I attempted to use the definition of an inverse
$$f(g(x))=x$$
so $$f'(g(x))g'(x)=1$$
so $$\frac{1}{\sqrt{1+g^4(x)}}g'(x)=1$$
which gives us $g'(x)=\sqrt{ 1+g^4(x)}$
beyond this I'm stuck.
could someone guide me on what I could do next?
 A: Now you have (after the edit(s)) that $\displaystyle f(0)=\int_0^0\frac{dt}{\sqrt{1+t^4}}=0$ so $f^{-1}(0)=g(0)=0$ so $g’(0)=\sqrt{1+(g(0))^4}=1.$
A: From what you have, $g'(0) = \sqrt{1+g^4(0)}$.
Now, $g(0) = f^{-1}(0)$, and by definition, $f(g(0)) = 0$. Note that $f(x) = 0$ only when $x=0$ since the integrated function is strictly positive. Hence, $g(0) = 0$ and $g'(0) = 1$.
A: We have $f(x)=\int_0^{x} \frac{dt}{\sqrt{1+t^4}}$, So by Lebnitz we have $f'(x)=\frac{1}{\sqrt{1+x^4}}$
Let $y=f(x)\implies x=f^{-1} (y)=g(y).$
Start with $f(g(x))=x$ or $g(f(x))=x$
D.w.r.t. $x$ to get $g'(f(x)) f'(x)=1 \implies g'(y_0)=\frac{1}{f'(x_0)}$
Where $(x_0,y_0)$ falls on the curve $y=f(x)$. Here, we have $x_0=0=y_0$, then
$$g'(0)=\frac{1}{f'(0)}=1$$
One may also write $$\frac{df^{-1}(y)}{dy}|_{y=y_0}=\frac{1}{f'(x_0)}$$
A: This is what you have to do next. Let $x=0$ in
$$f'(g(x))g'(x)=1.$$
Then you have
$$ f'(g(0))g'(0)=1. $$
Noting $g(0)=0$ and $f'(0)=\frac1{\sqrt{1+x^4}}\bigg|_{x=0}=1$,
you have
$$ g'(0)=1. $$
