# Question 232e8

Dec 12, 2015

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

#### Explanation:

Given $f \left(x\right) = {\left(\cos x\right)}^{2} / \left(1 + \sin x\right)$
*Note ${\left(\cos x\right)}^{2} = {\cos}^{2} x$

You can find the derivative using the quotient rule
$f \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$ then
 "f'(x) = (g(x)*h'(x) -h(x)*g'(x))/[g(x)]^2$\text{ " " } \left(1\right)$

1. Let $h \left(x\right) = {\left(\cos x\right)}^{2}$ then

Note: to differentiate $h \left(x\right)$ we need to use the power and chain rule

$h ' \left(x\right) = 2 {\left(\cos x\right)}^{2 - 1} \cdot 1 \left(- \sin x\right)$
$\textcolor{red}{h ' \left(x\right) = - 2 \sin x \cos x}$

1. Let $g \left(x\right) = 1 + \sin x$
$\textcolor{b l u e}{g ' \left(x\right) = \cos x \cdot 1} \implies \textcolor{b l u e}{\cos x}$

2. Let's substitute into (1) to get
$f ' \left(x\right) = \frac{\left(1 + \sin x\right) \textcolor{red}{\left(- 2 \sin x \cos x\right)} - {\cos}^{2} x \cdot \textcolor{b l u e}{\cos x}}{1 + \sin x} ^ 2$

Factot out the greatest common factor "-cos x#

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

Rewrite ${\cos}^{2} x = 1 - {\sin}^{2} x$ using Trigonometric Pythagorean identity

$= \frac{- \cos x \left(2 \sin x + 2 {\sin}^{2} x + \textcolor{g r e e n}{1 - {\sin}^{2} x}\right)}{1 + \sin x} ^ 2$

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

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

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

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

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

I hope this help.