If \( y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right), \quad -\frac{1}{\sqrt{3}} \leq x \leq \frac{1}{\sqrt{3}} \) Then, \( \frac{dy}{dx} \) is
Step-by-step Solution:
Put \(\tan \theta = x\) Then, \[ y = \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) = \tan^{-1}(\tan 3\theta) = 3\theta \] Now, \[ \frac{dy}{dx} = \frac{dy}{d\theta} \times \frac{d\theta}{dx} = 3 \times \frac{1}{\sec^2 \theta} = \frac{3}{1 + x^2} \]