Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question

y^{1/m} = x + \sqrt{1 + x^2}

, then

(1 + x^2)y_2 + x y_1

is equal to:

m^2 y

correct

m y^2

m^2 y^2

m y

Solution

Step 1: First derivative With

y = \left(x + \sqrt{1 + x^2}\right)^m

y_1 = m\left(x + \sqrt{1 + x^2}\right)^{m-1}\left(1 + \frac{x}{\sqrt{1 + x^2}}\right) = \frac{m\left(x + \sqrt{1 + x^2}\right)^m}{\sqrt{1 + x^2}} = \frac{my}{\sqrt{1 + x^2}}

Step 2: Square and clear the radical

\sqrt{1 + x^2}\, y_1 = my \implies (1 + x^2)y_1^2 = m^2 y^2

Step 3: Differentiate and divide by

2y_1

2x y_1^2 + 2(1 + x^2)y_1 y_2 = 2m^2 y y_1 \implies (1 + x^2)y_2 + x y_1 = m^2 y

Answer: (1)

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