# Question #7951d

Oct 11, 2016

$\frac{\tan x - x \ln \left(2 x\right) {\sec}^{2} x}{x {\tan}^{2} x}$

#### Explanation:

You just need to make use of the quotient rule and the chain rule.

Quotient Rule:

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

Chain Rule:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
(basically, you get the derivative of $f ' \left(x\right)$ then multiply by the derivative of $g \left(x\right)$)

Solution:

$\frac{d}{\mathrm{dx}} \left[\ln \frac{2 x}{\tan} x\right]$

By quotient rule:

$\left[1\right] \text{ } = \frac{\tan x \cdot {D}_{x} \left[\ln \left(2 x\right)\right] - \ln \left(2 x\right) \cdot {D}_{x} \left[\tan x\right]}{\tan} ^ 2 x$

The derivative of $\tan x$ is ${\sec}^{2} x :$

$\left[2\right] \text{ } = \frac{\tan x \cdot {D}_{x} \left[\ln \left(2 x\right)\right] - \ln \left(2 x\right) \cdot {\sec}^{2} x}{\tan} ^ 2 x$

To get the derivative of $\ln \left(2 x\right)$, you use chain rule. The derivative of $\ln \left(x\right)$ is $\frac{1}{x}$

$\left[3\right] \text{ } = \frac{\tan x \cdot \frac{1}{2 x} \cdot {D}_{x} \left[2 x\right] - \ln \left(2 x\right) \cdot {\sec}^{2} x}{\tan} ^ 2 x$

$\left[4\right] \text{ } = \frac{\tan x \cdot \frac{1}{2 x} \cdot 2 - \ln \left(2 x\right) \cdot {\sec}^{2} x}{\tan} ^ 2 x$

$\left[5\right] \text{ } = \frac{\tan x \cdot \frac{1}{x} - \ln \left(2 x\right) \cdot {\sec}^{2} x}{\tan} ^ 2 x \cdot \frac{x}{x}$

$\left[6\right] \text{ } = \textcolor{red}{\frac{\tan x - x \ln \left(2 x\right) {\sec}^{2} x}{x {\tan}^{2} x}}$