Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question

x + \cos\theta = \sec\theta

and

y + \cos^8\theta = \sec^8\theta

, then

\left(\dfrac{x^2 + 4}{y^2 + 4}\right)\left(\dfrac{dy}{dx}\right)^2

is:

8

16

64

correct

49

Solution

Step 1: Differentiate parametrically With

x = \sec\theta - \cos\theta

and

y = \sec^8\theta - \cos^8\theta

\frac{dy}{dx} = \frac{8\sec^8\theta\tan\theta + 8\cos^7\theta\sin\theta}{\sec\theta\tan\theta + \sin\theta} = \frac{8\tan\theta(\sec^8\theta + \cos^8\theta)}{\tan\theta(\sec\theta + \cos\theta)} = \frac{8(\sec^8\theta + \cos^8\theta)}{\sec\theta + \cos\theta}

Step 2: Express in terms of

x

and

y

Since

\sec\theta\cos\theta = 1

x^2 + 4 = (\sec\theta - \cos\theta)^2 + 4 = (\sec\theta + \cos\theta)^2

y^2 + 4 = (\sec^8\theta - \cos^8\theta)^2 + 4 = (\sec^8\theta + \cos^8\theta)^2

Step 3: Form the required quantity

\left(\frac{dy}{dx}\right)^2 = \frac{64(\sec^8\theta + \cos^8\theta)^2}{(\sec\theta + \cos\theta)^2} = \frac{64(y^2 + 4)}{x^2 + 4}

\left(\frac{x^2 + 4}{y^2 + 4}\right)\left(\frac{dy}{dx}\right)^2 = 64

Answer: (3)

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