alleryIf $x^y = y^x$, then $\dfrac{dy}{dx}$ equals:
A$\dfrac{\ln y - \ln x}{x - y}$
B$1$
C$\dfrac{y(x \ln y - y)}{x(y \ln x - x)}$
D$y/x$
Answer & Solution
Correct answer: C. $\dfrac{y(x \ln y - y)}{x(y \ln x - x)}$
Take $\ln$: $y \ln x = x \ln y$.
**Differentiate:** $y'\ln x + y/x = \ln y + (x/y)y'$.
Solve for $y'$: $y'(\ln x - x/y) = \ln y - y/x$
⇒ $y' = (y(x\ln y - y))/(x(y\ln x - x))$.
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