d/dx(tan^-1sqrtx) =কত? 

ধরি, \(y = \tan^{-1}(\sqrt{x})\)

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

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

এখানে, \(u = \sqrt{x}\) ধরে, চেইন রুল ব্যবহার করে পাই,

\(\frac{dy}{dx} = \frac{d}{du} (\tan^{-1}u) \cdot \frac{du}{dx}\)

\(= \frac{1}{1+u^2} \cdot \frac{d}{dx}(\sqrt{x})\)

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

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

\(= \frac{1}{2\sqrt{x}(1+x)}\)

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