# Question 48c60

Apr 13, 2017

$- \frac{2 {\sec}^{2} x}{1 + \tan x} ^ 2$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$\text{Given " f(x)=(g(x))/(h(x))" then}$

• f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}} \frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$

$\text{here } g \left(x\right) = 1 - \tan x \Rightarrow g ' \left(x\right) = - {\sec}^{2} x$

$\text{and } h \left(x\right) = 1 + \tan x \Rightarrow h ' \left(x\right) = {\sec}^{2} x$

$\Rightarrow f ' \left(x\right) = \frac{\left(1 + \tan x\right) \left(- {\sec}^{2} x\right) - \left(1 - \tan x\right) \left({\sec}^{2} x\right)}{1 + \tan x} ^ 2$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{- {\sec}^{2} x \cancel{- {\sec}^{2} x \tan x} - {\sec}^{2} x \cancel{+ {\sec}^{2} x \tan x}}{1 + \tan x} ^ 2$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = - \frac{2 {\sec}^{2} x}{1 + \tan x} ^ 2$

Apr 13, 2017

$- \frac{2}{\cos x + \sin x} ^ 2 ,$
or,
$- \frac{2}{1 + \sin 2 x} .$

#### Explanation:

Observe that, $\frac{1 - \tan x}{1 + \tan x} = \tan \left(\frac{\pi}{4} - x\right) .$

Hence, d/dx{(1-tanx)/(1+tanx)}=d/dx{tan(pi/4-x),

=sec^2(pi/4-x)*d/dx(pi/4-x),...[because," The Chain Rule]"#

$= - {\sec}^{2} \left(\frac{\pi}{4} - x\right) ,$

$= - \frac{1}{\cos \left(\frac{\pi}{4} - x\right)} ^ 2 ,$

$- \frac{1}{\cos \left(\frac{\pi}{4}\right) \cos x + \sin \left(\frac{\pi}{4}\right) \sin x} ^ 2 ,$

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

$= - \frac{2}{\cos x + \sin x} ^ 2 , \mathmr{and} ,$

$= - \frac{2}{{\cos}^{2} x + 2 \cos x \sin x + {\sin}^{2} x} ,$

$= - \frac{2}{1 + \sin 2 x} .$

Enjot Maths.!