## Calculus 10th Edition

$1+\ln xy=e^{x-y}$ $0+\frac{1}{x}y+\ln(x)\frac{dy}{dx}=(1-\frac{dy}{dx})e^{x-y}$ $0+\frac{1}{x}y+\ln(x)\frac{dy}{dx}=e^{x-y}-e^{x-y}\frac{dy}{dx}$ $(e^{x-y}+\ln x)\frac{dy}{dx}=e^{x-y}-\frac{y}{x}$ $\frac{dy}{dx}=\frac{e^{x-y}-\frac{y}{x}}{(e^{x-y}+\ln x)}$ at point (1,1), $\frac{dy}{dx}=\frac{e^{1-1}-\frac{1}{1}}{(e^{1-1}+\ln1)}$ $=\frac{e^0-1}{e^0+0}$ $=0$ Equation of tangent: y-1=0(x-1) y-1=0 y=1