How do you differentiate #ln x^(1/3)#?

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Answer 1 · 2016-07-29T13:39:30

#1/(3x)#

There are two methods: one that simplifies the function first, and the other which doesn't.

Without simplifying the function:

#y=ln(x^(1/3))#

We will need to use the chain rule. Since:

#d/dxln(x)=1/x#

The chain rule tells us that:

#d/dxln(u)=1/u*(du)/dx#

Thus:

#dy/dx=1/x^(1/3)*d/dxx^(1/3)#

Use the power rule to find #d/dxx^(1/3)#:

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

Simplify:

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

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

#dy/dx=1/(3x)#

Alternatively, simplify first:

You may remember a rule of logarithms that states that:

#log(a^b)=b*log(a)#

Thus:

#ln(x^(1/3))=1/3*ln(x)#

So:

#y=1/3*ln(x)#

Now when differentiating, we won't have to use the chain rule, and the #1/3# is simply brought out of the differentiation:

#dy/dx=1/3*d/dxln(x)#

#dy/dx=1/3*1/x#

#dy/dx=1/(3x)#

Generalization:

This method can be applied to differentiating any function in the form:

#y=ln(x^a)#

Since #y=a*ln(x)#, we know that #dy/dx=a*1/x=a/x#.

This can also be proven using the power rule, but above way is simpler.