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

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Answer 1 · 2014-09-22T21:21:40

You much use the chain rule to solve this problem

#y=ln(x^2)#

#y'= 1/x^2*2x#

#y'= (2x)/x^2=2/x->#Solution

Or you could use one of the properties of logarithms where you can pull the power to the front of the expression as a factor.

#y=ln(x^2)=2ln(x)#

#y'= 2*1/x=2/x ->#Solution