Problem Statement
Find $\frac{dy}{dx}$ if $x^y + y^x = a^b$, where $a, b > 0$ are constants.
Verified Solution & Marking Scheme
Differentiate u = xʸ
$\ln u = y \ln x \implies \frac{du}{dx} = x^y \left(\frac{y}{x} + \ln x \frac{dy}{dx}\right)$.
Differentiate v = yˣ
$\ln v = x \ln y \implies \frac{dv}{dx} = y^x \left(\ln y + \frac{x}{y}\frac{dy}{dx}\right)$.
Combine and Solve for dy/dx
$\frac{du}{dx} + \frac{dv}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \ln y}{x^y \ln x + x y^{x-1}}$