x^x = y^x হলে dy/dx এর মান কত?

দেওয়া আছে, \(x^x = y^x\)

উভয় পক্ষে \(log\) নিয়ে পাই,

\(log(x^x) = log(y^x)\)

\(\implies xlogx = xylogy\)

এখন, \(x\) এর সাপেক্ষে অন্তরীকরণ করে পাই,

\(\frac{d}{dx}(xlogx) = \frac{d}{dx}(xylogy)\)

\(\implies x \cdot \frac{1}{x} + logx \cdot 1 = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \cdot logy + ylogy \cdot 1\)

\(\implies 1 + logx = \frac{xlogy}{y} \cdot \frac{dy}{dx} + ylogy\)

\(\implies \frac{xlogy}{y} \cdot \frac{dy}{dx} = 1 + logx - ylogy\)

\(\implies \frac{dy}{dx} = \frac{y(1 + logx - ylogy)}{xlogy}\)

যেহেতু \(x^x = y^x\), তাই \(xlogx = xylogy\), সুতরাং \(logx = logy + logx - 1\)

\(\implies logx = ylogy\)

অতএব,

\(\frac{dy}{dx} = \frac{y(1 + logx - ylogy)}{xlogy}\)

\(\implies \frac{dy}{dx} = \frac{y(1 + logx - logx)}{xlogy}\) [ যেহেতু ylogy = logx ]

\(\implies \frac{dy}{dx} = \frac{y}{xlogy}\)

আবার, \(xlogx = xylogy\) থেকে পাই, \(logy = \frac{logx}{y}\)

সুতরাং,

\(\frac{dy}{dx} = \frac{y}{x \cdot \frac{logx}{y}}\)

\(\implies \frac{dy}{dx} = \frac{y^2}{xlogx}\)

এখন প্রথম সমীকরণ থেকে পাই, \(logx = \frac{ylogy}{x}\)

সুতরাং,

\(\frac{dy}{dx} = \frac{y(1 + logx - ylogy)}{xlogy}\)

\(\implies \frac{dy}{dx} = \frac{y(1 + logx - ylogy)}{xlogy}\)

\(\implies \frac{dy}{dx} = \frac{y(1 + logx - x)}{xlogy}\)

আমরা জানি, \(ylogy = logx\) অথবা \(logx= \frac{xlogx}{x}\)

সুতরাং,

\(\frac{dy}{dx} = \frac{y(x \cdot \frac{xlogx}{x} +logx - \frac{xlogx}{x})}{logx}\)

\(\implies \frac{dy}{dx} = \frac{y/x (xlogx - y)}{ylogy-x}\) 🤔

\(\implies \frac{dy}{dx} = \frac{y}{x} \cdot \frac{(1+log x -ylogy)}{logy}\)