How do you find the derivative of #y=(lnx)^3#?

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Mar 21, 2018

Application of chain rule.
https://goo.gl/images/TN12uo

#dy/dx#=#3(lnx)^2 × 1/x#

[ First considering #lnx# as say p , it becomes #y=p^3# , then differentiating it, you'll get #dy/dx=3p^2# as #p^n=np^(n-1)# rule of differentiation,
Then,
Considering #x# of the #lnx# as #p# you are differentiating #lnp# which equals #(1/p)#, where#ln# is nothing but natural logarithm, #log base (e) (p)#
P.S.- I'm considering another variable to clear the idea.]

Hope it helps. :)

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