Question

If \( y = \tan^{-1} \left( \frac{4x}{1+5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right) \), then \( \frac{dy}{dx} = \):

• A. \( \frac{1}{25x^2 + 1} \)
• B. \( \frac{5}{1 + x^2} \)
• C. \( \frac{2}{1 + x^2} \)
• D. \( \frac{1}{1 + x^2} \)

Hint:

When differentiating inverse tangent functions, apply the chain rule and use the sum rule for the angles.

Answer 1

The Correct Option is D

Using the sum rule for inverse tangents and the chain rule for differentiation, we find that \( \frac{dy}{dx} = \frac{1}{1 + x^2} \).
Final Answer: \[ \boxed{\frac{1}{1 + x^2}} \]