# How do you differentiate ln((sin^2)x)?

Oct 30, 2016

Rewrite it first.

#### Explanation:

$\ln \left({\sin}^{2} x\right) = \ln \left({\left(\sin x\right)}^{2}\right) = 2 \ln \left(\sin x\right)$

Now use $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$, to get

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

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

$= 2 \left(\frac{1}{\sin} x\right) \left(\cos x\right)$

$= 2 \cot x$

Oct 30, 2016

$y = \ln \left({\left(\sin x\right)}^{2}\right)$

Which means that:

${e}^{y} = {\left(\sin x\right)}^{2}$

Use implicit differentiation on the left hand side of the equation and the chain rule on the right hand side of the equation:

${e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \sin x \cdot \cos x$

Divide expressions on both sides of the equation by ${e}^{y}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \sin x \cdot \cos x}{e} ^ y$

Don't forget that ${e}^{y}$ is ${\left(\sin x\right)}^{2}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \sin x \cos x}{\sin x \sin x}$

Simplify the fraction above:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \cos \frac{x}{\sin} x$

$\cos \frac{x}{\sin} x$ is $\cot x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cot x$

Presto!!