What is the derivative of # f(x)=(x^3)/(9 (3lnx-1))#?

1 Answer
mason m
Jan 28, 2016

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

Explanation:

First, note that this can be simplified as

#f(x)=x^3/(27lnx-9)#

To differentiate this, we can use the quotient rule, which states that

#d/dx[(g(x))/(h(x))]=(g'(x)h(x)-h'(x)g(x))/[h(x)]^2#

Applying this to the function at hand, we see that

#f'(x)=((27lnx-9)d/dx[x^3]-x^3d/dx[27lnx-9])/(27lnx-9)^2#

These derivatives are fairly simple to find. The one thing that may be tricky is remembering that #d/dx[lnx]=1/x#, so #d/dx[27lnx]=27/x#.

#f'(x)=((27lnx-9)(3x^2)-x^3(27/x))/(27lnx-9)^2#

Simplify.

#f'(x)=(81x^2lnx-27x^2-27x^2)/(27lnx-9)^2#

#f'(x)=(27x^2(3lnx-2))/(81(3lnx-1)^2)#

Note that while it appears a #9# was factored from the denominator, it really was #9^2# since the #9# was factored from two terms, since the denominator is squared.

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

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