y^x=e^x+y হলে dy/dx এর মান কোনটি?

Explanation:

Another Explanation (5): দেওয়া আছে, \(y^x = e^{x+y}\) উভয় পক্ষে \(ln\) নিয়ে পাই, \(ln(y^x) = ln(e^{x+y})\) \(\implies x \cdot ln(y) = (x+y) \cdot ln(e)\) \(\implies x \cdot ln(y) = x+y\) [ যেহেতু \(ln(e) = 1\)] এখন, \(x\) এর সাপেক্ষে অন্তরকলন করে পাই, \(\frac{d}{dx} (x \cdot ln(y)) = \frac{d}{dx} (x+y)\) \(\implies x \cdot \frac{d}{dx} (ln(y)) + ln(y) \cdot \frac{d}{dx} (x) = \frac{d}{dx} (x) + \frac{dy}{dx}\) \(\implies x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + ln(y) \cdot 1 = 1 + \frac{dy}{dx}\) \(\implies \frac{x}{y} \cdot \frac{dy}{dx} + ln(y) = 1 + \frac{dy}{dx}\) \(\implies \frac{x}{y} \cdot \frac{dy}{dx} - \frac{dy}{dx} = 1 - ln(y)\) \(\implies \frac{dy}{dx} (\frac{x}{y} - 1) = 1 - ln(y)\) \(\implies \frac{dy}{dx} (\frac{x-y}{y}) = 1 - ln(y)\) \(\implies \frac{dy}{dx} = \frac{y(1 - ln(y))}{x-y}\) সুতরাং, \(\frac{dy}{dx} = \frac{y(1 - ln(y))}{x-y}\) 🥳 অতএব, নির্ণেয় মান: \(\frac{y(1 - ln(y))}{x-y}\).