Question
$\text{If y}=\text{e}^{\text{y}}(\text{x}+1)=1,\text{ show that }\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
✓
Answer
Here $\text{e}^{\text{y}}(\text{x}+1)=1$
$\therefore\ \log[\text{e}^{\text{y}}(\text{x}+1)]=\log1$
$\therefore\ \log\text{e}^{\text{y}}(\text{x}+1)=0$
$\therefore\ \text{y}\log\text{e}=-\log(\text{x}+1)$
$\therefore\ \text{y}=-\log(\text{x}+1)\ \ [\because\text{loge}=1]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}+1}\ \dots(1)$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Bigg[\frac{(\text{x}+1).\frac{\text{d}}{\text{dx}}(1)-1.\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]$ $=-\Big[\frac{(\text{x}+1).0-1.1}{(\text{x}+1)^2}\Big]$
$=\frac{1}{(\text{x}+1)^2}=\Big(-\frac{1}{(\text{x}+1)^2}\Big)^2$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\ \dots[\because\text{of }(1)]$
$\therefore\ \log[\text{e}^{\text{y}}(\text{x}+1)]=\log1$
$\therefore\ \log\text{e}^{\text{y}}(\text{x}+1)=0$
$\therefore\ \text{y}\log\text{e}=-\log(\text{x}+1)$
$\therefore\ \text{y}=-\log(\text{x}+1)\ \ [\because\text{loge}=1]$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}+1}\ \dots(1)$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Bigg[\frac{(\text{x}+1).\frac{\text{d}}{\text{dx}}(1)-1.\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]$ $=-\Big[\frac{(\text{x}+1).0-1.1}{(\text{x}+1)^2}\Big]$
$=\frac{1}{(\text{x}+1)^2}=\Big(-\frac{1}{(\text{x}+1)^2}\Big)^2$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\ \dots[\because\text{of }(1)]$
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