Question

find dy/dx for y=sin^-1(1-x^2/1+x^2)​

Answer 1

Answer:

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Answer 2

$y = {sin}^{ - 1} ( \frac{1 - {x}^{2} }{1 + {x}^{2} } ) \\ \\ put \: \: x = tan \theta \\ \\ y = {sin}^{ - 1} ( \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } ) \\ \\ y = {sin}^{ - 1} (cos2 \theta) \\ \\ y= {sin}^{ - 1} (sin( \frac{\pi}{2} - 2 \theta)) \\ \\ y = \frac{\pi}{2} - 2\theta \\ \\ y = \frac{\pi}{2} - 2{tan}^{ - 1} x \\ \\ \frac{dy}{dx} = \frac{ - 2}{1 + {x}^{2} }$