y=tan^1(x/(sqrt(1-x^2))) হলে  dy/dx সমান-

প্রশ্ন: \(y = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\) হলে \(\frac{dy}{dx}\) সমান কত?

সমাধান:

ধরি, \(x = \sin\theta\). তাহলে, \(\theta = \sin^{-1}x\).

অতএব, \(y = \tan^{-1}\left(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\right)\)

\(\implies y = \tan^{-1}\left(\frac{\sin\theta}{\sqrt{\cos^2\theta}}\right)\)

\(\implies y = \tan^{-1}\left(\frac{\sin\theta}{\cos\theta}\right)\)

\(\implies y = \tan^{-1}(\tan\theta)\)

\(\implies y = \theta\)

\(\implies y = \sin^{-1}x\)

এখন, \(x\) এর সাপেক্ষে অন্তরকলন করে পাই,

\(\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}x)\)

\(\implies \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\)

সুতরাং, \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\). 🎉

উত্তর: \(\frac{1}{\sqrt{1-x^2}}\)