Question

If $e^x+ e^y=e^{x+y}$, prove that $\frac{dy}{dx}=-e^{y-x}$

Answer 1

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: {e}^{x} + {e}^{y} = {e}^{x + y}$

$\sf \: {e}^{x} + {e}^{y} = {e}^{x} \: {e}^{y}$

$\sf \: \dfrac{{e}^{x} + {e}^{y}}{{e}^{x} \: {e}^{y}} = 1$

$\sf \: \dfrac{{e}^{x}}{{e}^{x} \: {e}^{y}} + \dfrac{{e}^{y}}{{e}^{x} \: {e}^{y}} = 1$

$\sf \: {e}^{ - y} + {e}^{ - x} = 1$

On differentiating both sides w. r. t. x, we get

$\sf \: \dfrac{d}{dx}[ {e}^{ - y} + {e}^{ - x} ] = \dfrac{d}{dx}(1)$

$\sf \: \dfrac{d}{dx}{e}^{ - y} + \dfrac{d}{dx}{e}^{ - x} = 0$

$\sf \: {e}^{ - y}\dfrac{d}{dx}( - y) + {e}^{ - x}\dfrac{d}{dx}( - x) = 0$

$\sf \: - {e}^{ - y}\dfrac{dy}{dx} - {e}^{ - x} = 0$

$\sf \: - {e}^{ - y}\dfrac{dy}{dx} = {e}^{ - x}$

$\sf \: \dfrac{dy}{dx} = - \dfrac{{e}^{ - x}}{{e}^{ - y}}$

$\implies\sf \: \dfrac{dy}{dx} \: = \: - \: {e}^{y - x}$

$\rule{190pt}{2pt}$

Formulae Used

$\sf \: \dfrac{d}{dx}[ f(x) + g(x)] = \dfrac{d}{dx}f(x) + \dfrac{d}{dx}g(x)$

$\sf \: \dfrac{d}{dx}k = 0$

$\sf \: \dfrac{d}{dx}{e}^{x} = {e}^{x}$

$\sf \: \dfrac{d}{dx}x = 1$

$\sf \: {x}^{m} \div {x}^{n} = {x}^{m - n}$