Explanation: 
Another Explanation (5):
y = \(\tan^{-1}\sqrt{x}\) হলে, \(\frac{dy}{dx}\) নির্ণয়:
আমরা জানি, \(\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}\)
এখানে, y = \(\tan^{-1}\sqrt{x}\)
সুতরাং, \(\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}\sqrt{x})\)
= \(\frac{1}{1+(\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x})\) [চেইন রুল ব্যবহার করে]
= \(\frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}\) [\(\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\)]
= \(\frac{1}{2\sqrt{x}(1+x)}\)
অতএব, \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}(1+x)}\) 🥳🥳
