sin^-1x এর সাপেক্ষে tan^-1x এর অন্তরজ নির্ণয়

\(u = \tan^{-1}x\) এবং \(v = \sin^{-1}x\)

\(\frac{du}{dx} = \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}\) 🚀

\(\frac{dv}{dx} = \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}\) 🎯

\(\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{1}{1+x^2}}{\frac{1}{\sqrt{1-x^2}}} = \frac{\sqrt{1-x^2}}{1+x^2}\) 🎉

সুতরাং, উত্তর: \(\frac{\sqrt{1-x^2}}{1+x^2}\) ✅