$α ∈ R$, $p=-1/x^2$ and $f : R → R$,
with $f(x)=x^αe^p$, if $x>0$
and $f(x)=0$ if $x≤0$.
How to determine the derivate of the function in all of $x ∈ R$ where it exists.
(The task suggests that we should distinguish cases $α > 1$ ,$ α = 1$ und $α < 1$)
First, we can start by calculating $f'(x)$ for $x>0$ $$f'(x)=\alpha x^{\alpha-1}e^{-\frac{1}{x^2}}+x^\alpha \frac{2x}{x^4}e^{-\frac{1}{x^2}}=(\alpha-2p)x^{\alpha-1}e^p$$ and when $x=0$, we calculate $$\lim\limits_{x\rightarrow0^+}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\rightarrow0^+}x^{\alpha-1}e^{-\frac{1}{x^2}}\color{red}=0\quad\forall\alpha\in]0,+\infty]$$ $\color{red}=$ because the growth of exponential always wins against the polynomial one.
Thus, $f'(0)=\lim\limits_{x\rightarrow0^+}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\rightarrow0^-}\frac{f(x)-f(0)}{x-0}=0$
Now, for $\alpha=1$ we have $$f'(x)=\left\{ \begin{array}[cr] \ (1-2p)e^p&x>0\\ 0&x\leq 0 \end{array}\right.$$ for $\alpha\neq1$ we have $$f'(x)=\left\{ \begin{array}[cr] \ (\alpha-2p)x^{\alpha-1}e^p&x>0\\ 0&x\leq 0 \end{array}\right.$$