Question

what is the derivative of e^{x} ?

Answer 1

this is one of the basic derivatives \[\frac{\text d}{\text dx}(e^x) = e^x\]
are you asking about the method using limits?

Answer 2

The derivative of \[\huge e^(-x) \ is \ -e^(-x)\]. That's the correct method, doing parts gives an integral of\[\huge e^(-x) * \sin(2x),\] so use parts again with \[\huge dv = \sin(2x)\], leaving an \[\huge e^ (-x) * \cos(2x)\] integral, then take this to the other side of the equals, rearrange to make the integral the subject, leaving the answer as\[\huge e^(-x)/3 * (2\sin(2x) - \cos(2x))\]

Answer 3

also the answer is a e^{x} right @Master.RohanChakraborty

Answer 4

thanks !

Answer 5

\[
\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}\\
\lim_{h\to 0}\frac{e^xe^h-e^x}{h}\\
\lim_{h\to 0}\frac{e^x(e^h-1)}{h}
\]So our task is simply to prove that \[\lim_{h\to 0}\frac{(e^h-1)}{h}=1\]We can do this using deltas and epsilons. There might be an easier way, if so I can't think of it off the top of my head.

Answer 6

thanks all