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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
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Verified Answer
The correct answer is:
$-e^{-z}$
Given, $y=\frac{\log _e x}{x}$ and $z=\log _e x$
$\therefore \quad y=\frac{z}{e^z}$
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
$\begin{aligned}
\frac{d^2 y}{d z^2} & =\frac{e^z(-1)-(1-z) e^z}{e^{2 z}} \\
& =\frac{-2+z}{e^z}
\end{aligned}$
$\frac{d^2 y}{d z^2}+\frac{d y}{d z}=\frac{1-z}{e^z}+\frac{-2+z}{e^z}=-\frac{1}{e^z}=-e^{-z}$
$\therefore \quad y=\frac{z}{e^z}$
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
$\begin{aligned}
\frac{d^2 y}{d z^2} & =\frac{e^z(-1)-(1-z) e^z}{e^{2 z}} \\
& =\frac{-2+z}{e^z}
\end{aligned}$
$\frac{d^2 y}{d z^2}+\frac{d y}{d z}=\frac{1-z}{e^z}+\frac{-2+z}{e^z}=-\frac{1}{e^z}=-e^{-z}$
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