Question

Given, y = tan^-1 √(x^2 – 1) then what is the value of (2x^2 – 1)y’ + x(x^2 – 1)y”?

(a) -1

(b) 0

(c) 1

(d) 2

This question was posed to me in an interview for internship.

My question is taken from Second Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Right choice is (b) 0

Best explanation: We have, y = tan^-1 √(x^2 – 1)

Differentiating both sides with respect to x, we get,

y’ = 1/(1 + x^2 – 1)* d/dx(x^2 – 1)^1/2

1/x^2 * 1/2((x^2 – 1))^1/2 * 2x

Squaring both sides,

x^2(x^2 – 1)(y’)^2 = 1

Differentiating again with respect to x we get,

x^2(x^2 – 1)(y’)^2 + (y’)^2d/dx(x^4 – x^2) = d/dx(1)

Or, x^2(x^2 – 1)^2y’y” + (y’)^2(4x^3 – 2x) = 0

Or, x^2(x^2 – 1)y” + (y’)(2x^3 – x) = 0