# What is the derivative of f(x) = ln(sin^2(x))?

##### 1 Answer
Nov 25, 2015

$\frac{d}{\mathrm{dx}} \ln \left({\sin}^{2} \left(x\right)\right) = 2 \cot \left(x\right)$

#### Explanation:

We will use the following:
The chain rule
d/dxf(g(x)) = f'(g(x)g'(x)

$\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

$\frac{d}{\mathrm{dx}} \sin \left(x\right) = \cos \left(x\right)$

Now, as the function given is a logarithm of a power of the sine function, we will apply the chain rule twice:

$\frac{d}{\mathrm{dx}} \ln \left({\sin}^{2} \left(x\right)\right) = \frac{1}{{\sin}^{2} \left(x\right)} \cdot \left(\frac{d}{\mathrm{dx}} {\sin}^{2} \left(x\right)\right)$

$\implies \frac{d}{\mathrm{dx}} \ln \left({\sin}^{2} \left(x\right)\right) = \frac{1}{{\sin}^{2} \left(x\right)} \cdot 2 \sin \left(x\right) \cdot \left(\frac{d}{\mathrm{dx}} \sin \left(x\right)\right)$

$\implies \frac{d}{\mathrm{dx}} \ln \left({\sin}^{2} \left(x\right)\right) = \frac{2 \sin \left(x\right)}{{\sin}^{2} \left(x\right)} \cos \left(x\right) = \frac{2 \cos \left(x\right)}{\sin} \left(x\right) = 2 \cot \left(x\right)$