যদি y^x=x^y হয় তাহলে dy/dx এর মান কত হবে–

দেওয়া আছে, \(y^x = x^y\)

উভয় পক্ষে \(ln\) নিয়ে পাই,

\(ln(y^x) = ln(x^y)\)

\(\implies x \cdot ln(y) = y \cdot ln(x)\)

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

\(\frac{d}{dx} [x \cdot ln(y)] = \frac{d}{dx} [y \cdot ln(x)]\)

\(\implies x \cdot \frac{d}{dx} [ln(y)] + ln(y) \cdot \frac{d}{dx} [x] = y \cdot \frac{d}{dx} [ln(x)] + ln(x) \cdot \frac{d}{dx} [y]\)

\(\implies x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + ln(y) \cdot 1 = y \cdot \frac{1}{x} + ln(x) \cdot \frac{dy}{dx}\)

\(\implies \frac{x}{y} \cdot \frac{dy}{dx} + ln(y) = \frac{y}{x} + ln(x) \cdot \frac{dy}{dx}\)

\(\implies \frac{x}{y} \cdot \frac{dy}{dx} - ln(x) \cdot \frac{dy}{dx} = \frac{y}{x} - ln(y)\)

\(\implies \frac{dy}{dx} \cdot (\frac{x}{y} - ln(x)) = \frac{y}{x} - ln(y)\)

\(\implies \frac{dy}{dx} = \frac{\frac{y}{x} - ln(y)}{\frac{x}{y} - ln(x)}\)

\(\implies \frac{dy}{dx} = \frac{y/x - lny}{x/y - lnx}\)

\(\implies \frac{dy}{dx} = \frac{lny-y/x}{lnx-x/y}\) \(\times \frac{-1}{-1}\)

অতএব, \(\frac{dy}{dx} = \frac{lny - \frac{y}{x}}{lnx - \frac{x}{y}}\) 🎉