Question

The derivative of $ \frac{d}{dx} \left( \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) \right) \text{ is equal to} $

• A. \( \frac{3}{1 + x^2} \)
• B. \( \frac{3}{1 + 9x^2} \)
• C. \( \sec^2{3x} \)
• D. \( \frac{1}{1 + x^2} \)

Hint:

When differentiating composite functions like \( \tan^{-1}(u) \), apply the chain rule and simplify the expression for \( u \) carefully.

Answer 1

The Correct Option is B

Let \( y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) \). We need to differentiate this function using the chain rule. The derivative of \( \tan^{-1}(u) \) is: \[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] In our case, \( u = \frac{3x - x^3}{1 - 3x^2} \).
The derivative of \( u \) is: \[ \frac{du}{dx} = \frac{d}{dx} \left( \frac{3x - x^3}{1 - 3x^2} \right) \] By applying the quotient rule and simplifying, the result is: \[ \frac{3}{1 + 9x^2} \] Conclusion. Thus, the derivative is \( \frac{3}{1 + 9x^2} \), which corresponds to option (b).