How do you derive y = (1 + sin(2x)]/[1 - sin(2x)) using the quotient rule?

Jan 4, 2016

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

Explanation:

By the quotient rule, the derivative of $\frac{u \left(x\right)}{v \left(x\right)}$ is $\frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{v {\left(x\right)}^{2}}$. Let's say $f \left(x\right) = \frac{1 + \sin \left(2 x\right)}{1 - \sin \left(2 x\right)}$.

Here, $u \left(x\right) = 1 + \sin \left(2 x\right)$ and $v \left(x\right) = 1 - \sin \left(2 x\right)$. The derivative of $\sin \left(2 x\right)$ is $2 \cos \left(2 x\right)$ by the chain rule, so $u ' \left(x\right) = 2 \cos \left(2 x\right)$ and $v ' \left(x\right) = - u ' \left(x\right)$.

So, by applying the quotient rule :
$f ' \left(x\right) = \frac{2 \cos \left(2 x\right) \left(1 - \sin \left(2 x\right)\right) - \left(1 + \sin \left(2 x\right)\right) \left(- 2 \cos \left(2 x\right)\right)}{1 - \sin \left(2 x\right)} ^ 2$

$f ' \left(x\right) = \frac{2 \cos \left(2 x\right) \left(1 - \sin \left(2 x\right) + 1 + \sin \left(2 x\right)\right)}{1 - \sin \left(2 x\right)} ^ 2 = \frac{4 \cos \left(2 x\right)}{1 - \sin \left(2 x\right)} ^ 2$