Question

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

• A. $e^{-z}$
• B. $2 e^{-z}$
• 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
$\frac{d y}{d z}=\frac{e^{z}(1)-z e^{z}}{e^{2 z}}=\frac{1-z}{e^{z}}$
Again differentiating, we get
$\frac{d^{2} y}{d z^{2}} =\frac{e^{z}(-1)-(1-z) e^{z}}{e^{2 z}} $
$=\frac{-2+z}{e^{z}}$
$\therefore \frac{d^{2} y}{d z^{2}}+\frac{d y}{d z}=\frac{1-z}{e^{z}}+\frac{-2+z}{e^{z}}=-\frac{1}{e^{z}}=-e^{-z}$