Hello.

So I have this:

$\displaystyle x^{y} = y^{x}$

which, in case $\displaystyle x$ and $\displaystyle y$ are positive, may be graphed like:

Assuming:

$\displaystyle x^{y} = y^{x} \Longrightarrow e^{y \cdot ln(x)} = e^{x \cdot ln(y)} \Longrightarrow y \cdot ln(x) = x \cdot ln(y) \Longrightarrow ln(x^{y}) = ln(y^{x})$

It seems plotting $\displaystyle x^{y} = y^{x}$ and $\displaystyle ln(x^{y}) = ln(y^{x})$ should be essentially the same.

However, my real problem is when I try to get $\displaystyle \frac{dy}{dx}$ out of those equations.

*Unfortunately first pictures showing my thorough work had to be removed because of some reader-disruptive flaws.*

Essence of derivation of equation #1:

$\displaystyle \frac{d}{dx}(x^{y} = y^{x})$

$\displaystyle y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^{y} \cdot ln(x) \cdot \frac{dy}{dx} = x \cdot y^{x-1} \cdot \frac{dy}{dx} + y^{x} \cdot ln(y) \cdot \frac{dx}{dx} $

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

Essence of derivation of equation #2:

$\displaystyle \frac{d}{dx}(ln(x^{y}) = ln(y^{x}))$

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

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

Why is there a big difference?