# How do you differentiate g(x) = [1 + sin(2x)]/[1 - sin(2x)]  using the product rule?

Jan 9, 2016

In order to forcedly use the product rule, we'll rewrite it as $g \left(x\right) = \left(1 + \sin \left(2 x\right)\right) {\left(1 - \sin \left(2 x\right)\right)}^{-} 1$

#### Explanation:

Also, we'll need chain rule to differentiate the second term.

• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

• For the first term, we'll rename $u = 2 x$

• For the second term, we'll rename $v = \left(1 - \sin \left(w\right)\right)$ and $w = 2 x$ (but the logic follows the same way that of $u$, $v$, and so on)

• Product rule: $\left(a b\right) ' = a ' b + a b '$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = 2 \cos \left(2 x\right) {\left(1 - \sin \left(2 x\right)\right)}^{-} 1 + \left(1 + \sin \left(2 x\right)\right) - {v}^{-} 2 \left(- 2 \cos \left(2 x\right)\right)$

Substituting $v$ and rearranging negative exponents:

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{2 \cos \left(2 x\right)}{1 - \sin \left(2 x\right)} + \frac{2 \cos \left(2 x\right) \left(1 + \sin \left(2 x\right)\right)}{1 - \sin \left(2 x\right)} ^ 2$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{2 \cos \left(2 x\right) \left(1 - \sin \left(2 x\right)\right) + 2 \cos \left(2 x\right) \left(1 + \sin \left(2 x\right)\right)}{1 - \sin \left(2 x\right)} ^ 2$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{2 \cos \left(2 x\right) \left(1 \cancel{- \sin \left(2 x\right)} + 1 + \cancel{\sin \left(2 x\right)}\right)}{1 - \sin \left(2 x\right)} ^ 2$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{4 \cos \left(2 x\right)}{1 - \sin \left(2 x\right)} ^ 2$