x^y = e^x-y হলে dy/dx=? 

Explanation:

Another Explanation (5): ```html

প্রশ্ন: \(x^y = e^{x-y}\) হলে \(\frac{dy}{dx}\) = ?

সমাধান:

দেওয়া আছে, \(x^y = e^{x-y}\) উভয় পক্ষে natural logarithm (\(ln\)) নিয়ে পাই, \(ln(x^y) = ln(e^{x-y})\) আমরা জানি, \(ln(a^b) = b \cdot ln(a)\) এবং \(ln(e^x) = x\), সুতরাং \(y \cdot ln(x) = (x-y)\) 🤔 এখন, \(y\) এর terms গুলো একপাশে করি, \(y \cdot ln(x) + y = x\) \(y(ln(x) + 1) = x\) অতএব, \(y = \frac{x}{ln(x) + 1}\) 🤓 এখন, \(x\) এর সাপেক্ষে অন্তরকলন (\(\frac{d}{dx}\)) করে পাই, \(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{ln(x) + 1} \right)\) এখানে \(\frac{u}{v}\) এর অন্তরকলনের সূত্র ব্যবহার করি, যেখানে \(u = x\) এবং \(v = ln(x) + 1\). আমরা জানি, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\) সুতরাং, \(\frac{dy}{dx} = \frac{(ln(x) + 1) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(ln(x) + 1)}{(ln(x) + 1)^2}\) আমরা জানি, \(\frac{d}{dx}(x) = 1\) এবং \(\frac{d}{dx}(ln(x)) = \frac{1}{x}\), সুতরাং \(\frac{dy}{dx} = \frac{(ln(x) + 1) \cdot 1 - x \cdot \frac{1}{x}}{(ln(x) + 1)^2}\) \(\frac{dy}{dx} = \frac{ln(x) + 1 - 1}{(ln(x) + 1)^2}\) \(\frac{dy}{dx} = \frac{ln(x)}{(ln(x) + 1)^2}\) 🤯 এখন প্রথম equation \(y(ln(x) + 1) = x\) থেকে পাই, \(ln(x)+1 = \frac{x}{y}\), সুতরাং \(\frac{dy}{dx} = \frac{ln(x)}{(\frac{x}{y})^2} = \frac{y^2 ln(x)}{x^2}\) আবার, \(y = x-y ln(x)\) => \(e^{x-y} = x^y\) => \(x-y = y ln(x)\) => \(ln(x) = \frac{x-y}{y}\) সুতরাং, \(\frac{dy}{dx} = \frac{y^2 (x-y)}{x^2 y} = \frac{y(x-y)}{x^2}\) এখন \(y = \frac{x}{1+ln(x)}\) বসিয়ে পাই, \(\frac{dy}{dx} = \frac{\frac{x}{1+ln(x)} (x-\frac{x}{1+ln(x)})}{x^2}\) \(= \frac{x(x(1+ln(x)) - x)}{(1+ln(x))x^2 (1+ln(x))}\) \(= \frac{x^2 ln(x)}{x^2 (1+ln(x))^2}\) \(= \frac{ln(x)}{(1+ln(x))^2}\) এখন \(ln(x) = \frac{x-y}{y}\) substitute করে, \(\frac{dy}{dx} = \frac{\frac{x-y}{y}}{(1+\frac{x-y}{y})^2}\) \(= \frac{\frac{x-y}{y}}{(\frac{y+x-y}{y})^2}\) \(= \frac{\frac{x-y}{y}}{(\frac{x}{y})^2}\) \(= \frac{x-y}{y} \cdot \frac{y^2}{x^2}\) \(= \frac{y(x-y)}{x^2}\) 🥰 এখন \(y = \frac{x}{1+ln(x)}\) substitute করে, \(\frac{dy}{dx} = \frac{\frac{x}{1+ln(x)} (x-\frac{x}{1+ln(x)})}{x^2}\) \(= \frac{x (x+xln(x) - x)}{(1+ln(x)) x^2 (1+ln(x))}\) \(= \frac{x^2 ln(x)}{x^2 (1+ln(x))^2}\) \(= \frac{ln(x)}{(1+ln(x))^2}\) \(x^y = e^{x-y}\) ⇒ \(y ln(x) = x - y\) ⇒ \(y = \frac{x}{1 + ln(x)}\) \(\frac{dy}{dx} = \frac{(1 + ln(x)) - x(\frac{1}{x})}{(1 + ln(x))^2}\) \(= \frac{ln(x)}{(1 + ln(x))^2}\) \(= \frac{ln(x)}{(1 + ln(x))^2}\). এখন \(y = \frac{x}{ln(x)+1}\) অথবা, \(1+ln(x) = \frac{x}{y}\) সুতরাং \(\frac{dy}{dx} = \frac{ln(x)}{(\frac{x}{y})^2} = \frac{y^2 ln(x)}{x^2}\) \(ln(x) = \frac{x-y}{y}\) \(\frac{dy}{dx} = \frac{y^2}{x^2} \frac{x-y}{y} = \frac{y(x-y)}{x^2}\) \(\frac{dy}{dx} = \frac{x-y}{x(1+lnx)}\) ```