d/(dx) sin(2tan^-1sqrt((1-x)/(1+x)))=? 

Explanation:

Another Explanation (5): আর্থাৎ, \(\frac{d}{dx} \sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right)\) এর মান নির্ণয় করতে হবে। 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, \(\tan^{-1}\sqrt{\frac{1-x}{1+x}} = \tan^{-1}(\tan(\theta)) = \theta\) তাহলে, \(\sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right) = \sin(2\theta)\) যেহেতু \(x = \cos(2\theta)\), তাই \(\sin(2\theta) = \sqrt{1-\cos^2(2\theta)} = \sqrt{1-x^2}\) এখন, \(\frac{d}{dx} \sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right) = \frac{d}{dx} \sqrt{1-x^2}\) \(\frac{d}{dx} \sqrt{1-x^2} = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}\) অতএব, \(\frac{d}{dx} \sin\left(2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right) = \frac{-x}{\sqrt{1-x^2}}\) 🥳🎉