Question

viii.If ex+ey=ex+y,show that dy/dx=-ey-x​

Answer 1

Answer:

${e}^{x} + {e}^{y} = {e}^{x + y}$

Differentiate wrt x

${e}^{x} + {e}^{y} . \frac{dy}{dx} = {e}^{x + y} . (1+\frac{dy}{dx}) \\ \\ ( {e}^{x + y} - {e}^{y} ) \frac{dy}{dx} = {e}^{x+y} -e^x\\ use \: \: eq \: 1 \\ \\ {e}^{x} \frac{dy}{dx} = {e}^{x} (e^y-1) \implies \: \frac{dy}{dx} = e^y-1$

Use first equation again to get

$e^y=e^{x+y}-e^x = e^x(e^y-1)\\e^{y-x}=(e^y-1)$