Question

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

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

Answer 1

Answer: (D)

$-e^{-z}$

We have $z=\ln x\Rightarrow x=e^z$

$\therefore y=\frac{\ln e^z}{e^z}=z{e}^{-z}$

$\Rightarrow \frac{dy}{dz}=e^{-z}-z{e}^{-z}$ .....(1)

And $\frac{d^{2}y}{d{z}^2}=-e^{-z}-e^{-z}+z{e}^{-z}$ .....(2)

Now add above two equations.

$\Rightarrow \frac{d^{2}y}{d{z}^2}+\frac{dy}{dz}=-e^{-z}$