Question

If $x^y-y^x=1$ , then the value of $\frac{dy}{dx}$ is :

Answer 1

$\begin{array}{c}{x}^y-y^x=1\\ y_1=x^y\\ \log y_1=y\log x\\ \frac{d{y}^{\prime }}{dx}=x^y\left(\frac{y}{x}+\log x.\frac{dy}{dx}\right)\\ y_2=y^x\\ \log y_2=x\log y\\ \frac{d{y}_2}{dx}=y^x\left(\frac{x}{y}.\frac{dy}{dx}+\log y\right)\\ \therefore x^y\left(\frac{y}{x}+\log x.\frac{dy}{dx}\right)=y^x\left(\frac{x}{y}.\frac{dy}{dx}+\log y\right)\\ y.x^{4-1}+x^y\log x.\frac{dy}{dx}=x{y}^{x-1}\frac{dy}{dx}+y^x\log y\\ \left(x^y\log x-x{y}^{x-1}\right)\frac{dy}{dx}=y^x\log y-4x^{y-1}\\ \frac{dy}{dx}=\left(\frac{y^x\log y-y{x}^{y-1}}{x^y.\log x-x{y}^{x-1}}\right)\end{array}$