How do I find $g(x)$ from $f(g(x))$ and $f '(x)$?

Question

Given $f(g(x)) = x$, and $f '(x) = 1 + [f(x)]^2$, how do I find $g '(x)$?

I have tried, subbing $g(x)$ into $f'(x)$ and got that $f'(g(x)) = 1+x^2$ but I am not sure if I am doing the right thing. It seems like a very simple problem but I'm having trouble with it.

Answer 1

Since $$ \frac{df}{dx} = 1 + f^2 $$ we have $$ \int \frac{df}{1+f^2} = \int dx $$ Then $$ \arctan{f} = x + C \iff f = \tan \left( x + C \right) $$

From which we immediately get $$ g(x)=\arctan x-C $$ Thus $$ g'(x) =\dfrac{1}{1+x^2} $$

Answer 2

Note that $$[f(g(x))]'=f'(g(x))g'(x)=[x]'=1$$ By assumption $f'(g(x))=1+[f(g(x))]^2=1+x^2$ and hence $$f'(g(x))g'(x)=(1+x^2)g'(x)=1\implies g'(x)=\frac1{1+x^2}$$