x^y\ =\ y^x\এর ক্ষেত্রে $\frac{dy}{dx}\ =\ ?$

$x^y\ =\ y^x\ \Rightarrow y\ln x\ =\ x\ln y\ \Rightarrow\frac{y}{x}+\ln x\ \frac{dy}{dx}\ =\ \ln y\ +\frac{x}{y}\times\frac{dy}{dx}\ \Rightarrow\frac{dy}{dx}\ =\ \frac{\ln y\ -\ \frac{y}{x}}{lbnx\ -\ \frac{x}{y}\ }\ \therefore\frac{dy}{dx}\ =\ \frac{y\left(x\ln y\ -\ y\right)}{x\left(y\ln x\ -x\right)}$