What is the derivative of ##e^(lnx)##?

This is a classic problem in calculus, one that appears to require an exponential derivative but instead can be made incredibly easy using simple algebra and then differentiation.

In Algebra II, we learn that because ##e^x## and ##lnx## are inverses, ##e^lna=a## (the e and natural log offset each other).

Therefore, ##e^lnx=x##.

Now the derivative of ##x## is 1, and the problem is done.