Question

Given the equation: \[ y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2, \] find the expression: \[ (1+4x^2)^2 y'' - 16. \]

• A. \( 8x y' \)
• B. \( -8x(1 + 4x^2) y' \)
• C. \( 8x(1 + 4x^2) y' \)
• D. \( -8x y' \)

Hint:

For implicit differentiation, apply the chain rule carefully and differentiate twice if needed.

Answer 1

The Correct Option is B

Step 1: Differentiate both sides Given: \[ y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2. \] Differentiate with respect to \( x \), \[ \frac{dy}{dx} = 2 (\tan^{-1} 2x) \cdot \frac{1}{1 + (2x)^2} \cdot 2 + 2 (\cot^{-1} 2x) \cdot \frac{-1}{1 + (2x)^2} \cdot 2. \] Since, \[ \tan^{-1} a + \cot^{-1} a = \frac{\pi}{2}, \] \[ \frac{dy}{dx} = 0. \] Step 2: Differentiate again \[ \frac{d^2y}{dx^2} = -\frac{8x}{(1 + 4x^2)^2}. \] Multiplying by \( (1+4x^2)^2 \) and subtracting 16, \[ (1+4x^2)^2 y'' - 16 = -8x(1 + 4x^2) y'. \]