What is the derivative of the given function at all points where it exists?

[mathmari]

Hey!

Let $a\in \mathbb{R}$. Find the derivative of the function $f:\mathbb{R}\rightarrow \mathbb{R}$ $$f(x)=\left\{\begin{matrix}
x^ae^{-\frac{1}{x^2}} & \text{ if } x>0\\
0 & \text{ if } x\leq 0
\end{matrix}\right.$$
in all the points $x\in \mathbb{R}$, where it exists. So, first we have to show if and the function is continuous, right? (Wondering)

For $x>0$ and $x<0$ the function is continuous, so we have to check at $x=0$.
$\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}x^ae^{-\frac{1}{x^2}}=0$, for each $a$, or not?
$\lim_{x\rightarrow 0^-}f(x)=0$.
$f(0)=0$

So, the function is continuous at $x=0$.

Then for $x>0$ is a product of differentiable functions, so it is differentiable. And for $x<0$ it is also differentiable. Is this correct? (Wondering)

So, we have to check again at $x=0$.

$$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{x^ae^{-\frac{1}{x^2}} }{x}=\lim_{x\rightarrow 0}x^{a-1}e^{-\frac{1}{x^2}}=0 \text{ for each } a$$ Right? (Wondering)

So, at $x=0$ it is also differentiable with derivative $0$.

For $x>0$ the derivative is $ax^{a-1}e^{-\frac{1}{x^2}}+x^a\frac{2}{x^3}e^{-\frac{1}{x^2}}$.

For $x<0$ the derivative is $0$.

Is everything correct? (Wondering)

At the exercise is says to take cases for $a$ : $a=1$, $a>1$, $a<1$. But why? (Wondering)

 

[HallsofIvy]

I suspect that they are referring to your step "[tex]\lim_{x\to 0^+} x^{a- 1}e^{-\frac{1}{x^2}}= 0[/tex] for all a". That is true but how you prove it is true depend upon whether a< 1, a= 1, or a> 1. That is, they expect you to prove that step.

 

[mathmari]

Ah ok.

So, for $a=1$ we have $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{xe^{-\frac{1}{x^2}} }{x}=\lim_{x\rightarrow 0}e^{-\frac{1}{x^2}}=0$$

For $a>1$ we have $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{x^ae^{-\frac{1}{x^2}} }{x}=\lim_{x\rightarrow 0}x^{a-1}e^{-\frac{1}{x^2}}=0\cdot 0=0$$

For $a<1$ we have $$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{x^ae^{-\frac{1}{x^2}} }{x}=\lim_{x\rightarrow 0}x^{a-1}e^{-\frac{1}{x^2}}=\lim_{x\rightarrow 0}\frac{1}{x^{1-a}}e^{-\frac{1}{x^2}}$$ Is this equa to $0$ because the exponential function grows faster that $x^{1-a}$ ? (Wondering)

 

[HallsofIvy]

Yes, the exponential grows faster than any power of x.

 

[mathmari]

Yes, the exponential grows faster than any power of x.

Ok. Thank you very much! (Sun)