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

Apr 23, 2018

$\frac{2 \cos x}{1 - \sin x} ^ 2$

#### Explanation:

First, factor the numerator.

$\left(1 - \sin x\right) \left(1 + \sin x\right)$

Then cancel a factor from the bottom.

$\frac{\left(1 - \sin x\right) \left(1 + \sin x\right)}{\left(1 - \sin x\right) \left(1 - \sin x\right)}$
$\frac{1 + \sin x}{1 - \sin x}$

We know the quotient rule:
$\frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g} {\left(x\right)}^{2}$

So let's insert:
$\frac{\cos x \cdot \left(1 - \sin x\right) - \left(- \cos x\right) \cdot \left(1 + \sin x\right)}{{\left(1 - \sin x\right)}^{2}}$

We get in the numerator:
$\left(\cos x \cdot \left(1 - \sin x\right) + \cos x \cdot \left(1 + \sin x\right)\right)$

$\left(\cos x \left(1 - \sin x + 1 + \sin x\right)\right)$

$\left(\cos x \left(2\right)\right)$

Now in total:

$\frac{2 \cos x}{1 - \sin x} ^ 2$

And it can be left this way. There is no more useful algebra to do. I double checked this answer.