Question

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

• A. $e^x$
• B. $-\frac{e^x}{(1+e^x{)}^3}$
• C. $-\frac{e^x}{(1+e^x{)}^2}$
• D. $\frac{1}{(1+e^x{)}^2}$

Answer 1

The correct option is B $-\frac{e^x}{(1+e^x{)}^3}$

$y=x+e^x$

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

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

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

$\frac{d^{2}x}{d{y}^2}=\frac{(1+e^x)\frac{d}{dy}1-1}{(1+e^x{)}^2}\frac{d}{dy}(1+e^x)$

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

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

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