Question
If $e^{x+y} - x = 0$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
✓
Answer
Here,
$e^{x+y} - x = 0$
$e^{x+y} = x .....(i)$
Differentiating it with respect to x using chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^{\text{x}+\text{y}}\big)=\frac{\text{d}}{\text{dx}}(\text{x})$
$\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=1$
$\text{x}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=1$
[Using euqation (i)]
$1+\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}-1$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
$e^{x+y} - x = 0$
$e^{x+y} = x .....(i)$
Differentiating it with respect to x using chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{x}^{\text{x}+\text{y}}\big)=\frac{\text{d}}{\text{dx}}(\text{x})$
$\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=1$
$\text{x}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=1$
[Using euqation (i)]
$1+\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}-1$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\text{x}}{\text{x}}$
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