Compute the derivative from the definition of $f(x)=e^{-1/x^2}$ for $x\neq 0.$ Is $f'(x)$ continuous? 
Compute the derivative from the definition of $f(x)=e^{-1/x^2}$ for $x\neq 0.$ Is $f'(x)$ continuous?

I tried using the definition of continuity and I wasn't sure how to show formally that $\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=f'(x).$ Obviously, the derivative is $\dfrac{2}{x^3}e^{-1/x^2}.$
Here's my attempt.
So by the definition of the derivative, the derivative of $f(x)$ at a point $x$ is
$$\lim\limits_{h\to 0}\dfrac{e^{-1/(x+h)^2}-e^{-1/x^2}}{h}=e^{-1/x^2}\lim\limits_{h\to 0}\dfrac{e^{-1/(x+h)^2+1/x^2}-1}{h}\\
=e^{-1/x^2}\lim\limits_{h\to 0}\dfrac{e^{(2h+h^2)/(x^2(x+h)^2)}-1}{h},$$ but here I'm stuck. How can I get to the desired limit? Any help would be appreciated.
I know that $\lim\limits_{h\to 0} \dfrac{e^h-1}{h}=1.$ Also, we want to show that $$\lim\limits_{h\to 0}\dfrac{e^{(2hx+h^2)/(x^2(x+h)^2)}-1}{h}=\dfrac{2}{x^3}$$
I was thinking of using the Taylor series expansion for $e^x,$ but isn't there some other way? Thanks in advance!
 A: Okay you can use something similar to chain rule by this 
$$\lim_{h\rightarrow 0} \frac{ e^{-1/(x+h)^2 } - e^{-1/x^2}}{h} $$ Then Multiply by $$\frac{-1/(x+h)^2 - -1/x^2}{-1/(x+h)^2 - -1/x^2}$$
To get 
$$ \lim_{h\rightarrow 0} \frac{ e^{-1/(x+h)^2 } - e^{-1/x^2}}{-1/(x+h)^2 - (-1/x^2)} \frac{ -1/(x+h)^2 - (-1/x^2)}{h} $$ 
Now we can distribute the limit
$$\lim_{h\rightarrow 0} \frac{ e^{-1/(x+h)^2 } - e^{-1/x^2}}{-1/(x+h)^2 - (-1/x^2)} \lim_{h\rightarrow 0 } \frac{ -1/(x+h)^2 - (-1/x^2)}{h}$$
Now we need to solve the first limit let $u = -1/(x+h)^2+1/x^2  $ then $u \rightarrow 0 $ as $h \rightarrow 0$ Hence 
$$\lim_{u \rightarrow 0 } \frac{ e^{u- 1/x^2} - e^{-1/x^2}}{u } = \lim_{u \rightarrow 0 }  \frac{e^{-1/x^2}(e^u-1)}{u} = e^{-1/x^2}$$
For the second limit 
$$ \lim_{h\rightarrow 0 } \frac{ -1/(x+h)^2 - (-1/x^2)}{h}$$
$$\lim_{h \rightarrow 0} \frac{ -x^2 + ( x^2 + 2xh + h^2)}{h(x+h)^2x^2}=\lim_{h\rightarrow 0} \frac{2x + h}{(x+h)^2 x^2 } = \frac{2}{x^3}$$
The answer is the product of the two limits i.e. $e^{-1/x^2} \frac{2}{x^3}$ which is the derivative of $f(x)$ right?? .
So in general if you have a composition of functions and you need to find the derivative using the definition say something like this 
$$\lim \frac{f(g(x+h)) - f(g(x))}{h} = \lim \frac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \frac{g(x+h) - g(x) }{h} $$
Then for the first limit we use a substitution $u = g(x+h) -g(x)$ then as $h$ goes to zero then $u$ goes to $0$ then 
$$\lim \frac{f(u + g(x)) - f(g(x))}{u} $$ this is similar to $\lim \frac{f(u+r) - f(r)}{u}$ So in above you can make one substitution.  
A: Note that $$  \frac 1 {x^2} - \frac 1 {(x+h)^2} = \frac {h (2x + h)}{x^2 (x+h)^2} \xrightarrow {h \to 0} 0 \cdot \frac {2x}{x^4} =0,$$ so the difference of the two fractions is an infinitesimal. Now apply the limit you have learned: $$\frac {\mathrm e^h - 1}h \to 1 [h \to 0],$$
by substituting the infinitesimals to a simpler one according this limit expression. 
