How do you find $y'$ for $x^y = y^x$?

Question

Using the laws of logarithms: $y\ln(x) = x\ln(y)$, $y = x\frac{\ln(y)}{\ln(x)}$

Is it now quotient rule for the derivative? How is this done?

Answer 1

Note that:

$x^y=y^x \implies y\ln{x}=x\ln{y}\implies y'\ln{x}+\frac{y}{x} =\ln{y}+\displaystyle\frac{xy'}{y}$.

Isolating $y'$, we have

$$\displaystyle y'=\frac{\ln{y}-\frac{y}{x}}{\ln{x}-\frac{x}{y}}.$$

Answer 2

$$y\ln(x) = x\ln(y) $$

$$\Rightarrow y\cdot\frac{1}{x}+\ln(x)\cdot\frac{dy}{dx}=x\cdot\frac{1}{y}\cdot\frac{dy}{dx}+\ln(y)\cdot1$$

$$$$

Answer 3

You can try separating variables: $$x^y = y^x \Rightarrow \frac{y}{\ln y} = \frac{x}{\ln x}$$ Taking the derivative: $$\frac{\ln y - 1}{(\ln y)^2} dy= \frac{\ln x - 1}{(\ln x)^2} dx$$

$$\Rightarrow \frac{dy}{dx}= \frac{(\ln y)^2}{\ln y - 1} \frac{\ln x - 1}{(\ln x)^2}$$

Answer 4

Consider the implicit function $F(x,y)=x^y-y^x (=0)$. We have the derivative $y’=-\frac{F_x}{F_y}$. Hence $$y'= -\frac{yx^{y-1}-y^x\ln y}{x^y\ln x-xy^{x-1}}=$$ $$\frac{y^x \ln y-yx^{y-1}}{x^y\ln x-xy^{x-1}}=\frac{y^x}{x^y}\cdot \frac{\ln y-x^{y-1}/y^{x-1}}{\ln x-y^{x-1}/x^{y-1}}=\frac{\ln y-\frac yx}{\ln x-\frac xy}$$ where one has put $\frac{y^x}{x^y}=1$ and $\frac{x^{y-1}}{y^{x-1}}=\frac yx$