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

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Answer 1 · 2017-02-18T06:05:54

#dy/dx = 3((ln(x))^2/x)#

We can apply Chain Rule of Differentiation.

#d/dx f@g(x)= [d/dx f(u)][d/dx u]#

or

#[f@g(x)]' = {f'[g(x)]}{g'(x)}#

#Let# #u = lnx # and #f(x) = (...)^3#

#d/dx f@g(x) =[3(lnx)^2][d/dx lnx] #

#d/dx f@g(x) =[3(lnx)^2][1/x]#

#d/dx f@g(x) = 3((lnx)^2/x)#