(log x/ x)  এর অন্তরক সহগ কত? 

দেওয়া আছে, \(y = \log \frac{x}{x}\)

আমরা জানি, \(\frac{d}{dx} (\log x) = \frac{1}{x}\)

এবং \(\frac{d}{dx} (uv) = u\frac{dv}{dx} + v\frac{du}{dx}\)

তাহলে, \(\frac{dy}{dx} = \frac{d}{dx} \left(\frac{\log x}{x}\right)\)

এখানে, \(u = \log x\) এবং \(v = \frac{1}{x}\) ধরি।

সুতরাং, \(\frac{du}{dx} = \frac{1}{x}\) এবং \(\frac{dv}{dx} = -\frac{1}{x^2}\)

এখন, \(\frac{dy}{dx} = \log x \cdot \left(-\frac{1}{x^2}\right) + \frac{1}{x} \cdot \frac{1}{x}\)

\(= -\frac{\log x}{x^2} + \frac{1}{x^2}\)

\(= \frac{1 - \log x}{x^2}\)

অতএব, \(\frac{d}{dx} \left(\frac{\log x}{x}\right) = \frac{1 - \log x}{x^2}\) 🥳