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

Apr 30, 2018

the answer $f ' \left(x\right) = - \sin x$

#### Explanation:

The common way is

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

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

$f ' \left(x\right) = - \sin \left(x\right)$

By using the quotient rule

$y = f \frac{x}{g} \left(x\right)$

$y ' = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

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

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

$= \left(\frac{\sin \left(x\right) \cdot \left(1 - \sin {\left(x\right)}^{2}\right)}{\cos} {\left(x\right)}^{2}\right) - 2 \cdot \sin \left(x\right)$

= $\frac{\sin \left(x\right) \cdot {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right) - 2 \sin x$

= $\sin x - 2 \sin x$

= $- \sin x$