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

Jan 28, 2016

$f ' \left(x\right) = \frac{{x}^{2} \left(3 \ln x - 2\right)}{3 {\left(3 \ln x - 1\right)}^{2}}$

#### Explanation:

First, note that this can be simplified as

$f \left(x\right) = {x}^{3} / \left(27 \ln x - 9\right)$

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

$\frac{d}{\mathrm{dx}} \left[\frac{g \left(x\right)}{h \left(x\right)}\right] = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{h \left(x\right)} ^ 2$

Applying this to the function at hand, we see that

$f ' \left(x\right) = \frac{\left(27 \ln x - 9\right) \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] - {x}^{3} \frac{d}{\mathrm{dx}} \left[27 \ln x - 9\right]}{27 \ln x - 9} ^ 2$

These derivatives are fairly simple to find. The one thing that may be tricky is remembering that $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$, so $\frac{d}{\mathrm{dx}} \left[27 \ln x\right] = \frac{27}{x}$.

$f ' \left(x\right) = \frac{\left(27 \ln x - 9\right) \left(3 {x}^{2}\right) - {x}^{3} \left(\frac{27}{x}\right)}{27 \ln x - 9} ^ 2$

Simplify.

$f ' \left(x\right) = \frac{81 {x}^{2} \ln x - 27 {x}^{2} - 27 {x}^{2}}{27 \ln x - 9} ^ 2$

$f ' \left(x\right) = \frac{27 {x}^{2} \left(3 \ln x - 2\right)}{81 {\left(3 \ln x - 1\right)}^{2}}$

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.

$f ' \left(x\right) = \frac{{x}^{2} \left(3 \ln x - 2\right)}{3 {\left(3 \ln x - 1\right)}^{2}}$