Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

u=f(\tan x)

v=g(\sec x)

and

f'(x)=\tan^{-1}x

g'(x)=\csc^{-1}x

, then

\dfrac{du}{dv}

x=\dfrac{\pi}{4}

\sqrt{2}

correct

2\sqrt{2}

3\sqrt{2}

4\sqrt{2}

Solution

Step 1: Compute

\dfrac{du}{dx}

and

\dfrac{dv}{dx}

\dfrac{du}{dx}=f'(\tan x)\cdot\sec^{2}x=\tan^{-1}(\tan x)\cdot\sec^{2}x=x\sec^{2}x

\dfrac{dv}{dx}=g'(\sec x)\cdot\sec x\tan x=\csc^{-1}(\sec x)\cdot\sec x\tan x

Step 2: Note

\csc^{-1}(\sec x)=\csc^{-1}\!\left(\dfrac{1}{\cos x}\right)=\sin^{-1}(\cos x)=\dfrac{\pi}{2}-x

. So

\dfrac{dv}{dx}=\left(\dfrac{\pi}{2}-x\right)\sec x\tan x

. Step 3:

\dfrac{du}{dv}=\dfrac{x\sec^{2}x}{\left(\dfrac{\pi}{2}-x\right)\sec x\tan x}=\dfrac{x\sec x}{\left(\dfrac{\pi}{2}-x\right)\tan x}

Step 4: At

x=\dfrac{\pi}{4}

\sec\dfrac{\pi}{4}=\sqrt{2}

\tan\dfrac{\pi}{4}=1

\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}

\dfrac{du}{dv}\bigg|_{x=\pi/4}=\dfrac{\dfrac{\pi}{4}\cdot\sqrt{2}}{\dfrac{\pi}{4}\cdot 1}=\sqrt{2}

Correct answer: (1)

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