How do you find the derivative of #x^(2x)#?

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Answer 1 · 2015-03-01T15:29:03

The answer is: #y'=2e^(2xlnx)(lnx+1)#.

Instead of remembering a complicate formula, we use these logarithmic properties:

#a=e^lna# or, better: #a^b=e^ln(a^b)=e^(blna)#.

So our function becomes:

#y=e^(2xlnx)#

and

#y'=e^(2xlnx)(2*1*lnx+2x*1/x)=2e^(2xlnx)(lnx+1)#.