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is the derivative of e^x is just e^x? and if i did the derivative of e^x again, nothing change?
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yes, it is \(e^x\) all the way down
is that simple XD?
are you trying to find the series for \(e^x\) ?
yeaup haha unless u get a e^2x then u have to pull the 2 out in front like a chain rule
it is a well known series \[\sum_{k=0}^{\infty}\frac{x^k}{k!}\]
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and yeah i was trying to use the Maclaurin series.
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