## Show a function is differentiable everywhere, and show its derivative is continuous

1. The problem statement, all variables and given/known data
show

$$f(x)=\left\{e^{\frac{-1}{x}} \\\ x>0$$
$$f(x)=\left\{0 \\\ x\leq 0$$
is differentiable everywhere, and show its derivative is continuous

2. Relevant equations
Product Rule and Chain Rule for derivatives. Definition of a derivative
$$f^{'}(a)=\frac{f(x)-f(a)}{x-a}$$

3. The attempt at a solution
Showing that f(x) is differentiable when x > 0 is easy. I use product rule and chain rule to define functions that when composed make f(x). Then for the derivative of f(x) when x>0, I show f'(x) is continuous by finding f''(x) which implies f'(x) is continuous.

I run into problems when x$$\leq$$0.

I say on x$$\leq$$0, $$f^{'}(o)=\frac{f(x)-f(o)}{x-o}=\frac{0}{x}=0$$ as x$$\rightarrow0$$ so f'(0) exists.
So now f'(x)=0 for all x < 0, and f'(x)$$\rightarrow$$0 as x$$\rightarrow$$0.

I run into my problem when trying to show that the derivative here is continuous. I'm not sure where to go from here. I'm really blanking and haven't really even made an attempt at a solution. My only idea is to maybe use right-hand and left-hand limits to show that the limit of f'(x) at 0 exists, so therefore it is also continuous at 0. I really feel I'm reaching here, and I'd appreciate some guidance.

Thanks.
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 You're right in that you need to show that the limit as x-->0 of f'(x) is f'(0) in order to show that the derivative is continuous. In order to do this you need to find the derivative...clearly it is 0 for x$$\leq$$0....what about for x>0? Differentiate, then show that the derivative approaches 0 as x-->0+. It should be clear that the derivative is continuous for x$$\neq$$0 and left-continuous at x=0, so all you need to do is show right continuity at 0.