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What is the derivative of #y = ln(2x^3-x^2)#?

1 Answer

Oct 8, 2016

#d/(dx)lnf(x)=(f'(x))/f(x)#

Explanation:

Recall that #d/(dx)lnx=1/x#

#y=ln(2x^3-x^2)# is a function within a function; we can apply chain rule.

Let #u=2x^3-x^2#, then #y=lnu#
#(dy)/(du)=1/u# and #(du)/(dx)=6x^2-2x#
#(dy)/(dx)=(dy)/(du)#x#(du)/(dx)=1/(2x^3-x^2)*(6x^2-2x)=(6x^2-2x)/(2x^3-x^2)#

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