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How do you find the first and second derivative of #lnx^2 #?

1 Answer

Oct 10, 2016

#y'=2x/x^2#

#y''=2/x^2#

Explanation:

Using the chain rule we get:

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

#y''=((x^2*2)-4x^2)/x^4#=#(2x^2-4x^2)/x^4#=#-2x^2/x^4#=#2/x^2#

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