Exercise: Show that $f:[0, \infty) \rightarrow [0,\infty), \ f(x) := x^2\ln(1+x^2)$ is bijective.
My approach:
Injectivity: Show that $\mathrm{kernel}(f)=0$.
Surjectivity: I tried to find an inverse function at this point because the following exercise would be to find the derivative of $f^{-1}$ at $\ln(2)$.
I tried the usual approach, namely:
$$\begin{align} y &= x^2 \ln(1+x^2) &\leadsto \\ \sqrt{y} &= x \sqrt{\ln(1+x^2)} &\leadsto \\ x &= \frac{\sqrt{y}}{\sqrt{\ln(1+x^2)}} ,\end{align}$$
but that didn't help.
I then looked at the inverse functions of both partial functions $x^2$ and $\ln(1+x^2)$ which are $\sqrt{x}=:r$ and $\sqrt{e^x-1}=:s$ because I figured $f^{-1}$ might be $\frac{s}{q}$ or $s \cdot q$, but that turned out to be wrong too.
So how do I find the inverse function? Is there even such an $f^{-1}$ that you can express in terms of elementary functions (going into flawr's answer) or would it simply be better to show the bijectivity without an inverse function? Thanks in advance!
To compute the derivative of the inverse we do not necessarily have to have a formula for the inverse itself:
Note that since $t = f(f^{-1}(t))$ by differentiating (assuming $f^{-1}$ exists and $f$ and $f^{-1}$ are differentiable) we get
$$ 1 = f'(f^{-1}(t)) \cdot [f^{-1}]'(t)$$
Therefore
$$ [f^{-1}]'(t) = \frac{1}{f'(f^{-1}(t))} $$
In your case $f(1) = \ln(2)$ and therefore $f^{-1}(\ln(2)) = 1$.
Therefore
$$ [f^{-1}]'(\ln(2)) = \frac{1}{f'(f^{-1}(\ln(2)))} = \frac{1}{f'(1)}$$
Note that at no point we had to compute $f^{-1}$.