y=tan^-1((4x)/(1-4x^2)) হলে dy/dx সমান কত?

প্রশ্ন: \( y=\tan^{-1}\left(\frac{4x}{1-4x^2}\right) \) হলে \( \frac{dy}{dx} \) নির্ণয় করো। 🤔

সমাধান:

ধরি, \( 2x = \tan{\theta} \)। 🤓

তাহলে, \( \theta = \tan^{-1}(2x) \) হবে। 👍

এখন, \( y = \tan^{-1}\left(\frac{4x}{1-4x^2}\right) \)

\( = \tan^{-1}\left(\frac{2(2x)}{1-(2x)^2}\right) \)

\( = \tan^{-1}\left(\frac{2\tan{\theta}}{1-\tan^2{\theta}}\right) \) 🥰

আমরা জানি, \( \frac{2\tan{\theta}}{1-\tan^2{\theta}} = \tan{2\theta} \)। 🤩

সুতরাং, \( y = \tan^{-1}(\tan{2\theta}) = 2\theta \)। 🥳

যেহেতু \( \theta = \tan^{-1}(2x) \), সুতরাং \( y = 2\tan^{-1}(2x) \) । 😎

এখন, \( \frac{dy}{dx} = \frac{d}{dx}\left(2\tan^{-1}(2x)\right) \)

\( = 2 \cdot \frac{d}{dx}\left(\tan^{-1}(2x)\right) \)

আমরা জানি, \( \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2} \)। 🤔

সুতরাং, \( \frac{d}{dx}(\tan^{-1}(2x)) = \frac{1}{1+(2x)^2} \cdot \frac{d}{dx}(2x) \)

\( = \frac{1}{1+4x^2} \cdot 2 \)

\( = \frac{2}{1+4x^2} \) । 😉

অতএব, \( \frac{dy}{dx} = 2 \cdot \frac{2}{1+4x^2} = \frac{4}{1+4x^2} \) । 🎉

সুতরাং, \( \frac{dy}{dx} = \frac{4}{1+4x^2} \)। ✅