\[ y = \sin^{-1} \left( \frac{x + \frac{1}{x}}{\sqrt{x^2 + 3 + \frac{1}{x^2}}} \right) = \sin^{-1} \left( \frac{x + \frac{1}{x}}{\sqrt{\left( x + \frac{1}{x} \right)^2 + 1}} \right) \] Let \[ x + \frac{1}{x} = \tan \theta \Rightarrow \sec^2 \theta \frac{d\theta}{dx} = 1 - \frac{1}{x^2} \] Also, \[ y = \sin^{-1} \left( \frac{\tan \theta}{\sec \theta} \right) = \theta \] So, \[ \frac{dy}{dx} = \frac{dy}{d\theta} \times \frac{d\theta}{dx} = 1 \times \frac{1 - \frac{1}{x^2}}{\sec^2 \theta} = \frac{1 - \frac{1}{x^2}}{\left( x + \frac{1}{x} \right)^2 + 1} \] \[ = \frac{x^2 - 1}{x^4 + 3x^2 + 1} \]