# How do you differentiate (ln(2x) )/ (cos2x) using the quotient rule?

Jan 11, 2016

$f ' \left(x\right) = \frac{\cos \left(2 x\right) + 2 x \ln \left(2 x\right) \cdot \sin \left(2 x\right)}{x {\cos}^{2} \left(2 x\right)}$

#### Explanation:

$f \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$

with:

$h \left(x\right) = \ln \left(2 x\right) \implies h ' \left(x\right) = \frac{1}{\cancel{2} x} \cdot \cancel{2} = \frac{1}{x}$

$g \left(x\right) = \cos \left(2 x\right) \implies g ' \left(x\right) = - \sin \left(2 x\right) \cdot 2 = - 2 \sin \left(2 x\right)$

Using the Quotient Rule

$f ' \left(x\right) = \frac{h ' \left(x\right) \cdot g \left(x\right) - h \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

then:

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

$= \frac{\frac{1}{x} \cdot \cos \left(2 x\right) + 2 \cdot \ln \left(2 x\right) \cdot \sin \left(2 x\right)}{\cos} ^ 2 \left(2 x\right) =$

$\frac{\cos \left(2 x\right) + 2 x \ln \left(2 x\right) \cdot \sin \left(2 x\right)}{x {\cos}^{2} \left(2 x\right)}$