Question

Differentiate :-

$y^x=x^y$

Answer 1

we take natural logs of both sides and simplify:

$ln\left(x^y\right)=ln\left(y^x\right)\to yln\left(x\right)=xln\left(y\right)$

use implicit differentiation, taking derivatives of both sides with respect to x:

$y\prime ln\left(x\right)+y\cdot \frac{1}{x}=1\cdot ln\left(y\right)+x\cdot \frac{y\prime }{y}$

Solving for $y^{\prime }$

$y\prime ln\left(x\right)-\frac{xy\prime }{y}=ln\left(y\right)-\frac{y}{x}$

Implies that:

$y\prime =\frac{ln\left(y\right)-\frac{y}{x}}{ln\left(x\right)-\frac{x}{y}}=\frac{xyln\left(y\right)-y^2}{xyln\left(x\right)-x^2}$