Question

Differentiate \[ \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \] with respect to \[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right). \]

Hint:

Use trigonometric identities to simplify inverse trig expressions before differentiating.

Answer 1

Let \[ u = \tan^{-1} \left( \frac{2x}{1 - x^2} \right), \quad v = \sin^{-1} \left( \frac{2x}{1 + x^2} \right). \] Notice that \[ \tan^{-1} \left( \frac{2x}{1 - x^2} \right) = 2 \tan^{-1} x, \] since \[ \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}. \] Also, \[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 2 \sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right), \] but more simply, \[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{-1} x, \] because \[ \sin(2 \theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}, \] and \(\sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{-1} x\). Therefore, \[ u = 2 \tan^{-1} x, \quad v = 2 \tan^{-1} x. \] Hence, \[ \frac{du}{dv} = \frac{d(2 \tan^{-1} x)}{d(2 \tan^{-1} x)} = 1. \]
Final answer: \[ \boxed{1}. \]