How do you find the derivative of #(lnx)/x^(1/3)#?

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Answer 1 · 2016-12-19T12:46:16

#dy/dx=(3-lnx)/(3x^(4/3))#

#y = lnx/root3x = lnx/x^(1/3)#

Applying the quotient rule:

#dy/dx = (x^(1/3)* 1/x - lnx* 1/3x^(-2/3))/x^(2/3)#

#= (x^(-2/3) - 1/3lnx*x^(-2/3)) / x^(2/3)#

#= (x^(-2/3)(1-lnx/3))/x^(2/3)#

#=(1-lnx/3)/x^(4/3)#

#=(3-lnx)/(3x^(4/3))#