Question

ex+ey=ex+y prove that dy/dx=ex(ey-1)/ey(ex-1)​

Answer 1

$Answer \: \\ \\ e {}^{x} + e {}^{y} = e {}^{(x + y)} \\ \\ Differentiate \: \: Both \: sides\: with \: \\ respect \: to \: \: x \: we \: have \\ \\ \frac{d(e {}^{x} )}{dx} + \frac{d(e {}^{y} )}{dx} = \frac{d(e {}^{(x + y)} )}{dx} \\ \\ e {}^{x} + e {}^{y} \times \frac{dy}{dx} = e {}^{(x + y)} \times (1 + \frac{dy}{dx} ) \\ \\ e {}^{x} - e {}^{(x +y )} = e {}^{(x + y)} \: \: \: \frac{dy}{dx} - e {}^{y} \: \: \: \frac{dy}{dx} \\ \\ e {}^{x} - e {}^{(x + y)} = \frac{dy}{dx} \: \: (e {}^{(x + y)} - e {}^{y} ) \\ \\ \frac{dy}{dx} = \frac{e {}^{x} - e {}^{(x + y)} }{e {}^{(x + y)} - e {}^{y} } \\ \\ \frac{dy}{dx} = \frac{e {}^{x} - e {}^{x} \times e {}^{y} } {e {}^{x} \times e {}^{y} - e {}^{y} } \\ \\ \frac{dy}{dx} = \frac{e {}^{x} (1 - e {}^{y} )}{e {}^{y}(e {}^{x} - 1) } \\ \\ \\ therefore \: \: Differentiation \: of \: \: \\ \\ e {}^{x} + e {}^{y} = e {}^{(x + y)} is \\ \\ \frac{e {}^{x} (1 - e {}^{y} )}{e {}^{y}(e {}^{x} - 1)}$