Question

If $y =\frac{\sec x +\tan x}{\sec x - \tan x} $ , then $\frac{dy}{dx} = $

• A. $2 \sec x(\sec x + \tan x)$
• B. $2 \sec^2 x(\sec x + \tan x)^2$
• C. $2 \sec x(\sec x + \tan x)^2$
• D. $\sec x(\sec x + \tan x)^2$

Answer 1

Answer: (C)

$y =\left(\frac{\sec x +\tan x}{\sec x - \tan x}\right) = \frac{1+\sin x}{1-\sin x} $
$\Rightarrow y=\frac{\left(1+\sin x\right)^{2}}{1-\sin^{2} x} =\frac{1+\sin^{2} x +2 \sin x}{\cos^{2} x}$
$ \Rightarrow y =\sec^{2} x+\tan^{2} x +2 \tan x \sec x$
$ \Rightarrow y =\left(\sec x + \tan x\right)^{2}$
Now Differentiating (i) w.r.t. x, we get ... (i)
$\frac{dy}{dx} = 2\left( \sec x + \tan x \right). \left[\sec x \tan x + \sec^{2} x \right] $
$=2 \sec x \left(\sec x + \tan x\right)\left(\sec x + \tan x\right)$
$ = 2 \sec x \left(\sec x + \tan x\right)^{2}$