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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{e}^{\text{x}+\text{y}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^{\text{x}}(\text{e}^\text{y}-1)}{\text{e}^{\text{y}}(\text{e}^{\text{x}}-1)}$ or $\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{y}-\text{x}}=0$
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Answer

$\text{e}^\text{x}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow\text{e}^\text{x}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}+\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}(\text{e}^\text{y}-\text{e}^{\text{x}+\text{y}})=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}}{\text{x}^\text{y}-\text{e}^{\text{x}+\text{y}}}$
$=\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}({1-\text{e}}^\text{x})}$
$=-\frac{\text{e}^\text{x}(\text{e}^\text{y}-1)}{\text{e}^\text{y}(\text{e}^\text{x}-1)}$

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