Question

If \( y = \tan(3\tan^{-1} x) \), then: \[ (1 - 3x^2)\frac{d^2y}{dx^2} - 12x \frac{dy}{dx} = \]

• A. \( 6(x + y) \)
• B. \( 6(y - x) \)
• C. \( 6y \)
• D. \( -6x \)

Hint:

Implicit Differentiation of Composite Inverse Functions}
Use identity for \( \tan(3\tan^{-1} x) \)
Carefully apply quotient rule for both first and second derivatives
Substitute \( y \) back after derivation for simplification

Answer 1

The Correct Option is B

Let: \[ y = \tan(3\tan^{-1} x) \] We know: \[ \tan(3\tan^{-1} x) = \frac{3x - x^3}{1 - 3x^2} \Rightarrow y = \frac{3x - x^3}{1 - 3x^2} \] Let’s denote \( y = \frac{u(x)}{v(x)} \), where: - \( u(x) = 3x - x^3 \) - \( v(x) = 1 - 3x^2 \) Compute first derivative using quotient rule: \[ \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} \] Then compute second derivative \( \frac{d^2y}{dx^2} \), plug into: \[ (1 - 3x^2)\frac{d^2y}{dx^2} - 12x \frac{dy}{dx} \] After simplification, we get: \[ = 6(y - x) \]