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ধরি, \(y = \tan^{-1}{\left ( \frac{a + b x}{b - a x} \right )}\)

আমরা জানি, \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}{\left ( \frac{x - y}{1 + x y} \right )}\)

প্রদত্ত রাশিটিকে আমরা লিখতে পারি:

\(y = \tan^{-1}{\left ( \frac{\frac{a}{b} + x}{1 - \frac{a}{b} x} \right )}\)

এটি \(\tan^{-1}A + \tan^{-1}B = \tan^{-1}{\left ( \frac{A + B}{1 - A B} \right )}\) এর অনুরূপ, যেখানে \(A = \frac{a}{b}\) এবং \(B = x\)।

সুতরাং, \(y = \tan^{-1}{\left ( \frac{a}{b} \right )} + \tan^{-1}x\)

\(x\) এর সাপেক্ষে অন্তরীকরণ করে পাই:

\(\frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}{\left ( \frac{a}{b} \right )} + \tan^{-1}x \right)\)

\(\frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}{\left ( \frac{a}{b} \right )} \right) + \frac{d}{dx} \left( \tan^{-1}x \right)\)

যেহেতু \(\frac{a}{b}\) একটি ধ্রুবক, তাই \(\frac{d}{dx} \left( \tan^{-1}{\left ( \frac{a}{b} \right )} \right) = 0\)।

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

সুতরাং, \(\frac{dy}{dx} = 0 + \frac{1}{1 + x^2}\)

\(\frac{dy}{dx} = \frac{1}{1 + x^2}\)

অন্তরক নির্ণয় কর : ( tan^ - 1 left ( a + b x b - a x right ) )