Question
If $e^y(x + 1) = 1$, then show that $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^{2}.$

Answer

$\text{e}^{\text{y}.} \text{(x + 1)} = 1 \Rightarrow \text{e}^{\text{y}}. \text{1 + (x + 1)}.\text{e}^{\text{y}}.\frac{\text{dy}}{\text{d}} = 0$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = -\frac{1}{\text{(x + 1)}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = + \frac{1}{\text{(x + 1)}^{2}} = \big(\frac{\text{dy}}{\text{dx}}\big)^{2}$

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