Show that the function $f(x)=e^{-\frac{1}{x}},\text{ if }x>0\wedge f(x)=0,\text{ if }x\leq 0$ is smooth and all derivatives vanish in $x=0$ I had an idea but I think it is wrong. I have said $e^z$ is smooth because the n-th derivative of $e^z$ is always $e^z$ and then I have substituted $z$ with $-\frac{1}{x}$. But just calculating the first derivative Shows me that I am probably wrong because:
$(e^{z})^{'}=e^{z}=e^{-\frac{1}{x}}$
But 
$(e^{z})^{'}=(e^{-\frac{1}{x}})^{'}=\frac{e^{-\frac{1}{x}}}{x^2}$
Then also 
$e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x^2}\iff 1=\frac{1}{x^2}$
Which is not true.
Why does the Substitution not work?
I have also tried to see a pattern in the derivatives to give an explicit Formula for the n-th derivative.
Using the product-formula for derrivatives
$$(e^{-\frac{1}{x}})^{'}=(e^{-\frac{1}{x}})(-\frac{1}{x})^{'}\overset{'}{\rightarrow}(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(-\frac{1}{x})^{''}=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(-\frac{2}{x^3})=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(\frac{1}{x^2})(-\frac{2}{x})=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})^{'}(-\frac{2}{x})=(e^{-\frac{1}{x}})^{'}((-\frac{1}{x})^{'}+(-\frac{2}{x}))\rightarrow ...$$
I have got the hunch that the n-th derivative can be written in a recursive form like
$$(e^{-\frac{1}{x}})^{(n)}\text{[ It is the derrivatove not the nth power ]}=(e^{-\frac{1}{x}})^{(n-1)}\cdot r$$
$r$ is the remainder and I am trying to figure out how it would look like. From my Observations so far I guess that it Looks like a sum $\sum_{k=1}^{n}(-1)^k\frac{n-k}{x^k}$. But I Need some help to prove it. 
My attempts so far have proved to be not very fruitful that's why I am asking for your support.
In order to Show that the derivative vanishes at 0 I first of all Need to find out what the derivative for positive $x$ Looks like. Then the plan would be to Show that the Limit ffrom the left as well as from the Right side is $0$
 A: The problem with your substitution is that the function $g(x)=1/x$ "blows up" at $x=0.$ You can avoid this problem by calculating the derivative directly. In the Wiki page cited in the comments, they use the Maclaurin expansion for $e^x$. You can also find the derivative using L'Hospital's Rule (to get the penultimate equality):
$\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac {e^{-1/x}}x = \lim_{x \to 0^+} \frac {1/x}{e^{1/x}} = \lim_{x \to 0^+} \frac 1 {e^{1/x}} = 0$
and then of course, we have, using the definition of $f$, 
$\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac {0}x =0 $
so we conclude that $f'(0)=0.$
Now it's not hard to show by induction that all derivatives vanish at $x=0.$
A: When $x<0$ then $f^{(n)}(x)=0$ for all $n\geq0$. When $x>0$ then $f^{(n)}(x)=p_n(1/x)e^{-1/x}$ for some polynomial $t\mapsto p_n(t)$. The latter is true for $n=0$ with $p_0(t)\equiv1$, and
$$f^{(n+1)}(x)={d\over dx}\biggl(p_n(1/x)e^{-1/x}\biggr)=\bigg({-1\over x^2}{p_n}'(1/x)+p_n(1/x){1\over x^2}\biggr)e^{-1/x}=:p_{n+1}(1/x)e^{-1/x}$$
for $x>0$ and $n\geq0$. Finally I claim that $f^{(n)}(0)=0$ for all $n\geq0$, and that $f^{(n)}$ is continuous at $0$. For $n=0$ this is true by definition of $f$. Assume that it is true for some $n\geq0$. Then
$$\lim_{x\to0+}{f^{(n)}(x)-f^{(n)}(0)\over x}=\lim_{x\to0+}{1\over x} p_n(1/x)e^{-1/x}=0\ ,$$
and trivially $$\lim_{x\to0-}{f^{(n)}(x)-f^{(n)}(0)\over x}=0\ .$$
This proves that $f^{(n+1)}(0)=0$.
A: First of all you look at the first derivative which is 
$\frac{e^{-\frac{1}{x}}}{x^2}$
Then you make the induction hypothesis that the $n$-th derivative of the function $f$ can be described as:
$e^{-\frac{1}{x}}\cdot P(x)$
Where $P(x)$ is even more General than a polynomial it is a finite sum  which can be described like $\sum_{k=z}^{m}a_kx^k$ where $z,m\in \mathbb{Z}$
for $n=1$ the induction hypothesis holds with $P(x)=x^{-2}=\frac{1}{x^2}$
Now you want to calculate 
$f^{(n+1)}$
Since you know that $f^{(n)}= e^{-\frac{1}{x}}\cdot P_n(x)$
You just take the derivative of the RHS
Applying the Product rule gives you
$e^{\frac{-1}{x}}\frac{P_n(x)}{x^2}+e^{-\frac{1}{x}}P_{n}'(x)= e^{-\frac{1}{x}}(\frac{P_n(x)}{x^2}+P_n'(x))$
To calculate the derivative of $P_n(x)$ you have to be careful since $(ax^0)'=0$ and not something like $-ax^{-1}$
And $\frac{P_n(x)}{x^2}+P_n'(x)=\sum_{k=z}^{m}a_kx^{k-2}+\sum_{k=z}^{-1}ka_kx^{k-1}+\sum_{k=1}^{m}ka_kx^{k-1}$ where $z,m\in \mathbb{Z}$
Because all three components are general polynomials the total sum must also be a general polynomial again which we can set as $P_{n+1}$. (Exc. to prove this!) 
Then you can Show that the Limit for $\lim_{x\downarrow 0} f^{(n)}(x)=0 ,\forall n\in\mathbb{N_0}$ by using
this result: Proof that $\lim_{x\downarrow 0}x^me^{\frac{-1}{x}} =0,m\in\mathbb{Z}$ with L'Hospital?
It does not make use of L'Hôspital tough it is similair to the solution presented in the comment what is different from the answer on wikivisual is that you don't need to know how the coefficients in the polynomial look live. Even if you would know the pattern in this case it doesn't make any difference.
