# If $f(x)=\sqrt{\large\frac{\sec x-1}{\sec x+1}}$$\: find\: f'(x). Also find ( f' \bigg( \large\frac{\pi}{2}\bigg).) ## 1 Answer Toolbox: • Whenever we see a function of the form y = \large \frac{u}{v}, which is a quotient of two other functions with derivatives, we can apply the following quotient rule: y'=\large \frac{1}{v^2}\;$$ ( v\large\frac{du}{dx}$$-u \large \frac{dv}{dx}$$)$
Step 1:
$f(x)=\sqrt{\large\frac{\sec x-1}{\sec x+1}}$
$\quad\;\;=\bigg(\large\frac{\sec x-1}{\sec x+1}\bigg)^{\Large\frac{1}{2}}$
$\therefore f'(x)$ can be found by using quotient rule $\large\frac{vu'-uv'}{v^2}$
Let $u=(\sec x-1)^{\Large\frac{1}{2}}$
$\therefore u'=\large\frac{1}{2}$$(\sec x-1)^{-\Large\frac{1}{2}}(\sec x\tan x) Let v=(\sec x-1)^{\Large\frac{1}{2}} \therefore v'=\large\frac{1}{2}$$(\sec x+1)^{-\Large\frac{1}{2}}(\sec x\tan x)$
Now applying this we get,
$f'(x)=\Large\frac{(\sec x+1)^{\Large\frac{1}{2}}.\Large\frac{1}{2(\sec x-1)^{\Large\frac{1}{2}}}-(\sec x-1)^{\Large\frac{1}{2}}.\large\frac{1}{2(\sec x+1)^{\Large\frac{1}{2}}}\times \sec x\tan x}{((\sec x+1)^{\Large\frac{1}{2}})^2}$
Step 2:
On simplifying we get,
$\qquad=\Large\frac{\sec x\tan x\bigg[\Large\frac{\sqrt{\sec x+1}}{2\sqrt{\sec x-1}}-\Large\frac{\sqrt{\sec x-1}}{2\sqrt{\sec x+1}}\bigg]}{\sec x+1}$
$\qquad=\Large\frac{\sec x\tan x\bigg[\Large\frac{\sec x+1-\sec x+1}{2\sqrt{\sec^2 x-1}}\bigg]}{(\sec x+1)}$
But $\sec^2x-1=\tan^2x$
$\therefore f'(x)=(\sec x\tan x)\bigg[\Large\frac{2}{2\tan x(\sec x+1)}\bigg]$
$\qquad\;\;\;=\large\frac{\sec x}{\sec x+1}$
Step 3:
$f'(\large\frac{\pi}{2})=\frac{\sec\Large\frac{\pi}{2}}{\sec \Large\frac{\pi}{2} +1}$
But $\sec (\Large\frac{\pi}{2})$$=0 \therefore f'(\large\frac{\pi}{2})$$=0$