- #1
roz77
- 16
- 0
So I know how to take the general derivative of this equation. It's a simple product rule. I have that. My problem is, I need to show that the derivative at x=0 is 0. I know that I'm supposed to use this equation.
f'(x)= lim x->0 of [f(x+h)-f(x)]/h
So I plug in x+h and I get:
[(e^(-1/(x+h)^2))-(e^(-1/x^2))]/h
My problem is, I have no idea how to simplify that. I know that I need to get rid of the denominator, but I'm not sure how I can do that. Suggestions?
f'(x)= lim x->0 of [f(x+h)-f(x)]/h
So I plug in x+h and I get:
[(e^(-1/(x+h)^2))-(e^(-1/x^2))]/h
My problem is, I have no idea how to simplify that. I know that I need to get rid of the denominator, but I'm not sure how I can do that. Suggestions?