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

Answer 1

Paisley Johnson

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

Use quotient rule:

#((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)#

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Answer 2

Scarlett Allison

To find the derivative of ln[x]/x^(1/3), you can use the quotient rule. The derivative of the quotient f(x)/g(x) is (f'(x)g(x) - f(x)g'(x)) / [g(x)]^2. Applying this rule to ln[x]/x^(1/3), where f(x) = ln[x] and g(x) = x^(1/3), and taking derivatives of f(x) and g(x) separately, you get:

f'(x) = 1/x and g'(x) = (1/3)x^(-2/3).

Substituting these into the quotient rule formula, you get:

[1/x * x^(1/3) - ln[x] * (1/3)x^(-2/3)] / [x^(1/3)]^2.

Simplify this expression to get the final derivative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.