How do you find the derivative of (ln(sin(2x)))^2?

1 Answer
Apr 15, 2016

$f ' \left(x\right) = 4 \ln \left(\sin \left(2 x\right)\right) \cdot \cot \left(2 x\right)$

Explanation:

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

let $u = \ln \left(\sin \left(2 x\right)\right)$
now $f \left(x\right) = {u}^{2}$

so $f ' \left(x\right) = 2 u \cdot \frac{\mathrm{du}}{\mathrm{dx}} = 2 \ln \left(\sin \left(2 x\right)\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

now to find $\frac{\mathrm{du}}{\mathrm{dx}}$

let $w = \sin \left(2 x\right)$
now $u = \ln \left(w\right)$

so $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{w} \cdot \frac{\mathrm{dw}}{\mathrm{dx}} = \frac{1}{\sin} \left(2 x\right) \cdot \frac{\mathrm{dw}}{\mathrm{dx}}$

now to find $\frac{\mathrm{dw}}{\mathrm{dx}}$

let $s = 2 x$
now $w = \sin \left(s\right)$

so $\frac{\mathrm{dw}}{\mathrm{dx}} = \cos \left(s\right) \cdot \frac{\mathrm{ds}}{\mathrm{dx}} = \cos \left(s\right) \cdot \frac{\mathrm{ds}}{\mathrm{dx}}$

We know that $\frac{\mathrm{ds}}{\mathrm{dx}} = 2$ so...

$\frac{\mathrm{dw}}{\mathrm{dx}} = \cos \left(2 x\right) \cdot 2$

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

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

simplify

$f ' \left(x\right) = 4 \ln \left(\sin \left(2 x\right)\right) \cdot \cos \frac{2 x}{\sin} \left(2 x\right)$

$f ' \left(x\right) = 4 \ln \left(\sin \left(2 x\right)\right) \cdot \cot \left(2 x\right)$