How do you differentiate #(sqrt(1+3x^2))lnx^2#?

1 Answer

May 1, 2015

Write it as:
#y=(1+3x^2)^(1/2)ln(x^2)#
and use the Product and Chain Rule:
#y'=1/2(1+3x^2)^(1/2-1)*(6x)ln(x^2)+(1+3x^2)^(1/2)1/x^2*2x=#
#=3xlnx^2/(sqrt(1+3x^2))+2sqrt(1+3x^2)/x=#
#=(3x^2lnx^2+2+6x^2)/(xsqrt(1+3x^2))#

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How do you differentiate #(sqrt(1+3x^2))lnx^2#?

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