If,y=x^(1/x)" "Then, dy/dx=? 

Explanation:

Another Explanation (5): দেওয়া আছে, \( y = x^{\frac{1}{x}} \) উভয় পক্ষে \( \ln \) নিয়ে পাই, \[ \ln y = \ln \left( x^{\frac{1}{x}} \right) \] \[ \ln y = \frac{1}{x} \ln x \] এখন, \( x \) এর সাপেক্ষে অন্তরকলন করে পাই, \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{x} \ln x \right) \] এখানে, গুণফল সূত্র ব্যবহার করে, \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \cdot \frac{d}{dx} (\ln x) + \ln x \cdot \frac{d}{dx} \left( \frac{1}{x} \right) \] \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \cdot \frac{1}{x} + \ln x \cdot \left( -\frac{1}{x^2} \right) \] \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^2} - \frac{\ln x}{x^2} \] \[ \frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \] \[ \frac{dy}{dx} = y \cdot \frac{1 - \ln x}{x^2} \] যেহেতু \( y = x^{\frac{1}{x}} \), তাই \[ \frac{dy}{dx} = x^{\frac{1}{x}} \cdot \frac{1 - \ln x}{x^2} \] \[ \frac{dy}{dx} = x^{\frac{1}{x} - 2} (1 - \ln x) \] সুতরাং, \( \frac{dy}{dx} = x^{\frac{1}{x} - 2} (1 - \ln x) \) 🥳🎉