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If $\mathrm{e}^{\mathrm{y}}(\mathrm{x}+1)=1$, show that $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2$
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$\mathrm{e}^{\mathrm{y}}(\mathrm{x}+1)=1 \Rightarrow \mathrm{e}^{\mathrm{y}}=\frac{1}{\mathrm{x}+1}$
$\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}+1)^2} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}+1)^2} \cdot \frac{1}{\mathrm{e}^{\mathrm{y}}}=\frac{-1}{(\mathrm{x}+1)}$
$\Rightarrow \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left(\frac{-1}{\mathrm{x}+1}\right)^2=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2$
$\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}+1)^2} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}+1)^2} \cdot \frac{1}{\mathrm{e}^{\mathrm{y}}}=\frac{-1}{(\mathrm{x}+1)}$
$\Rightarrow \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left(\frac{-1}{\mathrm{x}+1}\right)^2=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2$
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