# How do you differentiate f(x)= (1 + sin^2x)/(1 - sin2x)  using the quotient rule?

Sep 28, 2017

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

#### Explanation:

We must recall two things to make this simpler. First, $f \left(x\right) = g \frac{x}{h \left(x\right)} \to f ' \left(x\right) = \frac{h g ' - h ' g}{h} ^ 2$.

Here, $g \left(x\right) = 1 + {\sin}^{2} x = 1 + {\left(\sin x\right)}^{2}$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[1\right] + \frac{d}{\mathrm{dx}} \left[{\left(\sin x\right)}^{2}\right]$
$g ' \left(x\right) = 2 \cos x \sin x$

$h \left(x\right) = 1 - \sin \left(2 x\right)$
$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[1\right] - \frac{d}{\mathrm{dx}} \left[\sin 2 x\right]$
$h ' \left(x\right) = - 2 \cos 2 x$

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

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