If ex + xy = ex+y, prove that dy dx +e^ y-x =0 - Vidyadip

Question

If e^x + x^y = e^x+y, prove that $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$

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Answer

Here,
e^x + e^y = e^x+y ......(i)
Differentiating both the sides using chain rule,
$\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^\text{y})=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}+\text{y}})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{d}}{\text{dx}}\Big]$
$\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\Big(\frac{\text{e}^\text{x}+\text{e}^\text{x}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}-\text{e}^\text{y}}\Big)$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{y}-\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$

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