# Question #ce88a

Mar 10, 2017

$= \frac{1 - 2 \log x \ln \left(10\right)}{{x}^{3} \ln \left(10\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\log \frac{x}{x} ^ 2\right)$

This will require the quotient rule, the power rule, and knowing the derivitave of $\log x$.

Recall the quotient rule for finding a derivative:
$\frac{d}{\mathrm{dx}} \left(\frac{P \left(x\right)}{Q \left(x\right)}\right) = \frac{\left(Q \left(x\right) \cdot P ' \left(x\right)\right) - \left(P \left(x\right) \cdot Q ' \left(x\right)\right)}{Q {\left(x\right)}^{2}}$

Or simply, $\frac{B T ' - T B '}{B} ^ 2$ where $T$ is the top function and $B$ is the bottom function.

Also recall that $\frac{d}{\mathrm{dx}} {\log}_{a} \left(x\right) = \frac{1}{x \ln \left(a\right)}$ where $a$ is the base of the logarithm and $x$ is the argument. You won't use this very often but you just need to memorize this.

Let's proceed:

$\frac{d}{\mathrm{dx}} \left(\log \frac{x}{x} ^ 2\right)$

$= \frac{\left({x}^{2} \cdot \frac{1}{x \ln \left(10\right)}\right) - \left(\log x \cdot 2 x\right)}{{x}^{2}} ^ 2$

$= \frac{{x}^{2} / \left(x \ln \left(10\right)\right) - 2 x \log x}{x} ^ 4$

$= {x}^{2} / \left({x}^{4} \cdot x \ln \left(10\right)\right) - \frac{2 x \log x}{x} ^ 4$

$= \frac{1}{{x}^{3} \ln \left(10\right)} - \frac{2 \log x}{x} ^ 3$

Or

$= \frac{1 - 2 \log x \ln \left(10\right)}{{x}^{3} \ln \left(10\right)}$