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If $y=x+e^{x}$, then, what is $\frac{d^{2} x}{d y^{2}}$ equal to?
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Verified Answer
The correct answer is:
$-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$
Given that $y=x+e^{x}$ Differentiating w. r. t. $\mathrm{x}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=1+\mathrm{e}^{\mathrm{x}}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{1+\mathrm{e}^{\mathrm{x}}}$
Differentiating w. r.t. $\mathrm{y}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=-\frac{(1)\left(\mathrm{e}^{\mathrm{x}}\right)}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$
$=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{1}{\left(1+\mathrm{e}^{\mathrm{x}}\right)}=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=1+\mathrm{e}^{\mathrm{x}}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{1+\mathrm{e}^{\mathrm{x}}}$
Differentiating w. r.t. $\mathrm{y}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=-\frac{(1)\left(\mathrm{e}^{\mathrm{x}}\right)}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$
$=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{1}{\left(1+\mathrm{e}^{\mathrm{x}}\right)}=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$
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