Question

If $y=\frac{{\log }_{e}x}{x}$ and $z={\log }_{e}x$, then $\frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}$ is equal to

• A. $e^{-z}$
• B. $2e^{-z}$
• C. $z{e}^{-z}$
• D. $-e^{-z}$

Answer 1

Answer: (D)

Given, $y=\frac{{\log }_{e}x}{x}$ and $z={\log }_{e}x$
$\therefore y=\frac{z}{e^z}$
On differentiating w.r.t. $z$, we get
$\frac{dy}{dz}=\frac{e^z(1)-z{e}^z}{e^{2z}}=\frac{1-z}{e^z}$
Again differentiating, we get
$\frac{d^{2}y}{d{z}^2}=\frac{e^z(-1)-(1-z)e^z}{e^{2z}}$
$=\frac{-2+z}{e^z}$
$\therefore \frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}=\frac{1-z}{e^z}+\frac{-2+z}{e^z}=-\frac{1}{e^z}=-e^{-z}$