যদি y=sin{2tan^-1√(1-x)/(1+x)}  হয় তবে dy/dx=?

Explanation:

Another Explanation (5): দেওয়া আছে, \(y = \sin\left\{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\}\) ধরি, \(x = \cos{2\theta}\). তাহলে, \(\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}} = \sqrt{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}} = \sqrt{\tan^2{\theta}} = \tan{\theta}\) সুতরাং, \(y = \sin\{2\tan^{-1}(\tan{\theta})\} = \sin(2\theta)\) যেহেতু \(x = \cos{2\theta}\), তাই \(2\theta = \cos^{-1}{x}\) সুতরাং, \(y = \sin(\cos^{-1}{x})\) আমরা জানি, \(\sin(\cos^{-1}{x}) = \sqrt{1-x^2}\) অতএব, \(y = \sqrt{1-x^2}\) এখন, \(\frac{dy}{dx}\) নির্ণয় করি। \(\frac{dy}{dx} = \frac{d}{dx} (\sqrt{1-x^2}) = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}\) সুতরাং, \(\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}\) 🥳