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Mathematics Question on Continuity and differentiability

If y=sec⁡x+tan⁡xsec⁡x−tan⁡xy =\frac{\sec x +\tan x}{\sec x - \tan x} , then dydx=\frac{dy}{dx} =

A

2sec⁡x(sec⁡x+tan⁡x)2 \sec x(\sec x + \tan x)

B

2sec⁡2x(sec⁡x+tan⁡x)22 \sec^2 x(\sec x + \tan x)^2

C

2sec⁡x(sec⁡x+tan⁡x)22 \sec x(\sec x + \tan x)^2

D

sec⁡x(sec⁡x+tan⁡x)2\sec x(\sec x + \tan x)^2

Answer

2sec⁡x(sec⁡x+tan⁡x)22 \sec x(\sec x + \tan x)^2

Explanation

Solution

y=(sec⁡x+tan⁡xsec⁡x−tan⁡x)=1+sin⁡x1−sin⁡xy =\left(\frac{\sec x +\tan x}{\sec x - \tan x}\right) = \frac{1+\sin x}{1-\sin x}
⇒y=(1+sin⁡x)21−sin⁡2x=1+sin⁡2x+2sin⁡xcos⁡2x\Rightarrow y=\frac{\left(1+\sin x\right)^{2}}{1-\sin^{2} x} =\frac{1+\sin^{2} x +2 \sin x}{\cos^{2} x}
⇒y=sec⁡2x+tan⁡2x+2tan⁡xsec⁡x\Rightarrow y =\sec^{2} x+\tan^{2} x +2 \tan x \sec x
⇒y=(sec⁡x+tan⁡x)2\Rightarrow y =\left(\sec x + \tan x\right)^{2}
Now Differentiating (i) w.r.t. x, we get ... (i)
dydx=2(sec⁡x+tan⁡x).[sec⁡xtan⁡x+sec⁡2x]\frac{dy}{dx} = 2\left( \sec x + \tan x \right). \left[\sec x \tan x + \sec^{2} x \right]
=2sec⁡x(sec⁡x+tan⁡x)(sec⁡x+tan⁡x)=2 \sec x \left(\sec x + \tan x\right)\left(\sec x + \tan x\right)
=2sec⁡x(sec⁡x+tan⁡x)2= 2 \sec x \left(\sec x + \tan x\right)^{2}