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

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Answer 1 · 2017-04-14T19:13:45

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

#((ln(x))'(x^(1/3))-(x^(1/3))'(ln(x)))/(x^(1/3))^2#

Simplify:

#((1/x)(x^(1/3))-(1/3)(x^(-2/3))(ln(x)))/x^(2/3)#

Rewrite #(1/x)# as #x^-1# and merge it with #x^(1/3)#

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

Pull out a #x^(-2/3)# from the top:

#((x^(-2/3))(1-ln(x)/3))/x^(2/3)#

Take out the negative from #x^(-2/3)# and move it in the denominator to add exponents:

#(1-ln(x)/3)/(x^(4/3))#

To clean it up, multiply both the top and bottom by #3#:

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