Explanation:

Another Explanation (5):

প্রশ্ন: \( \frac{d}{dx} \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

সমাধান:

ধরি, \( x = \tan(\theta) \)। সুতরাং, \( \theta = \tan^{-1}(x) \)। তাহলে, \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \) আমরা জানি, \( \sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)} \) সুতরাং, \( \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta \) এখন, \( 2\theta = 2\tan^{-1}(x) \) অতএব, \( \frac{d}{dx} \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{d}{dx} \left(2\tan^{-1}(x)\right) \) আমরা জানি, \( \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \) সুতরাং, \( \frac{d}{dx} \left(2\tan^{-1}(x)\right) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \) সুতরাং, \( \frac{d}{dx} \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{2}{1+x^2} \) 🎉🎉🎉