Question

If y = x + e^x , then $\frac{d^{2}x}{d{y}^2}$ = ______________.

Answer 1

$y=x+e^x$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(x+e^x\right)$

$\Rightarrow \frac{dy}{dx}=1+e^x$

$\Rightarrow \frac{dx}{dy}=\frac{1}{1+e^x}\left(\frac{dx}{dy}=\frac{1}{{\displaystyle \frac{dy}{dx}}}\right)$

Differentiating both sides with respect to y, we get

$\frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dy}\left(\frac{1}{1+e^x}\right)$

$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{0-1\times {\displaystyle \frac{d}{dy}}\left(1+e^x\right)}{{\left(1+e^x\right)}^2}$

$\Rightarrow \frac{d^{2}x}{d{y}^2}=-\frac{0+e^x{\displaystyle \frac{dx}{dy}}}{{\left(1+e^x\right)}^2}$

$\Rightarrow \frac{d^{2}x}{d{y}^2}=-\frac{e^x{\displaystyle \left(\frac{1}{1+e^x}\right)}}{{\left(1+e^x\right)}^2}\left(\frac{dx}{dy}=\frac{1}{1+e^x}\right)$

$\Rightarrow \frac{d^{2}x}{d{y}^2}=-\frac{e^x}{{\left(1+e^x\right)}^3}$


If y = x + e^x , then $\frac{d^{2}x}{d{y}^2}$ = $\underline{-\frac{e^x}{{\left(1+e^x\right)}^3}}$.