If y = tan^-1 (x/(sqrt(1-x^2)))   then find the value of  (dy)/(dx)?  

দেওয়া আছে, \(y = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\).

ধরি, \(x = \sin\theta\). তাহলে, \(\theta = \sin^{-1}x\).

সুতরাং, \(y = \tan^{-1}\left(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\right) = \tan^{-1}\left(\frac{\sin\theta}{\cos\theta}\right) = \tan^{-1}(\tan\theta) = \theta\).

তাহলে, \(y = \theta = \sin^{-1}x\).

এখন, \(\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}\).

অতএব, \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\).

🎉🎉সুতরাং নির্ণেয় উত্তর: \(\frac{1}{\sqrt{1-x^2}}\)🎉🎉