What is the derivative of # (ln (x-4)) ^ 3#?

1 Answer
Ratnaker Mehta
Jul 3, 2016

The reqd. deri. #={3(ln(x-4))^2)/(x-4)#.

Explanation:

Let #y=(ln(x-4))^3#, & put #t=ln(x-4)#, so, #y=t^3............(1)#

So, #y# is a fun. of #t# & #t# of #x#.

By Chain Rule, then, we have, the reqd. deri. #=dy/dx=dy/dt*dt/dx#
#=d/dt(t^3)*d/dx{ln(x-4)}#..........[by #(1)#]

=#3t^2*{1/(x-4)}*d/dx(x-4)=(3t^2)/(x-4)*1={3(ln(x-4))^2)/(x-4)#.

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