If $g$ is the inverse of function $f$ and $f'(x)= \frac{1}{1+x^n}$, Find $g'(x)$

Question

I tried the question and got an answer by the following steps:

$f(g(x))=x$

Differentiating both sides w.r.t to $x$, we get

$f'(g(x)).g'(x)=1$

And therefore,

$g'(x)=1+\left[{g(x)} \right]^n$

Now my question is, are there any alternate ways to approach this problem? Especially ones which may need some Integration?

Answer 1

Original equation

$f'(x)= \frac{1}{1+x^n}$

Integrate it and define the result as $y$

$f(x)= \int \frac{1}{1+x^n}dx + C = y$

Invert $f(x)$ to get $g(y)$

$x = g(y)$

Differentiate w.r.t $y$, and use implicit differentiation to rewrite derivative via $\frac{dy}{dx}$ and compute it as a function of $x$. Then substitute $x = g(y)$.

$g'(y) = \frac{dx}{dy} = 1 / \frac{dy}{dx} = 1 + x^n = 1 + g(y)^n$

Essentially both my and your approaches do the same thing. I could technically integrate to get the explicit formulation $y(x)$, and then inverted it to substitute $x(y)$, thus getting an explicit expression for $g(y)$, but it was not necessary for an explicit expression. You will always integrate at some point, either solving for $y(x)$, or when solving the last ODE for $g'(y)$