# First principles differentiation question

tunabeast

## Homework Statement

Obtain the derivative of $$e^{2x}$$ from first principles.

## The Attempt at a Solution

I am upto the point where i have $$e^{2x}(\frac{e^{2h}-1}{h})$$ Where LIM h->0.

I just don't see how 2 is obtained from what's in the ().

## Answers and Replies

Science Advisor
Homework Helper
$$\frac{e^{2h}-1}{h}= 2\frac{e^{2h}-1}{2h}= 2\frac{e^u -1}{u}$$
if you let u= 2x.

Now, do you know what the limit, as u goes to 0 of
$$\frac{e^u- 1}{u}$$
is?

tunabeast
It tends to 1? I think i understand this now, thanks very much for your help.