Method of DifferentiationmediumFree

Method of Differentiation: Equal

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

Question
If y1/m=x+1+x2y^{1/m} = x + \sqrt{1 + x^2}, then (1+x2)y2+xy1(1 + x^2)y_2 + x y_1 is equal to:
Am2ym^2 ycorrect
Bmy2m y^2
Cm2y2m^2 y^2
Dmym y
Solution
Step 1: First derivative With y=(x+1+x2)my = \left(x + \sqrt{1 + x^2}\right)^m:
y1=m(x+1+x2)m−1(1+x1+x2)=m(x+1+x2)m1+x2=my1+x2y_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
1+x2 y1=my  ⟹  (1+x2)y12=m2y2\sqrt{1 + x^2}\, y_1 = my \implies (1 + x^2)y_1^2 = m^2 y^2
 Step 3: Differentiate and divide by 2y12y_1
2xy12+2(1+x2)y1y2=2m2yy1  ⟹  (1+x2)y2+xy1=m2y2x 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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