If y=frac{log _e x}{x} and z=log _e x, then f | vedclass

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(D) Given, $y=\frac{\log _e x}{x}$ and $z=\log _e x$. Since $z=\log _e x$, we have $x=e^z$. Substituting $x$ in $y$, we get $y=\frac{z}{e^z} = z e^{-z

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$-e^{-z}$

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(D) Given, $y=\frac{\log _e x}{x}$ and $z=\log _e x$. Since $z=\log _e x$, we have $x=e^z$. Substituting $x$ in $y$, we get $y=\frac{z}{e^z} = z e^{-z}$. Now, differentiating $y$ with respect to $z$: $\frac{d y}{d z} = \frac{d}{d z}(z e^{-z}) = e^{-z} + z(-e^{-z}) = e^{-z}(1-z)$. Again, differentiating with respect to $z$: $\frac{d^2 y}{d z^2} = \frac{d}{d z}(e^{-z}(1-z)) = -e^{-z}(1-z) + e^{-z}(-1) = e^{-z}(-1+z-1) = e^{-z}(z-2)$. Now, calculating $\frac{d^2 y}{d z^2} + \frac{d y}{d z}$: $\frac{d^2 y}{d z^2} + \frac{d y}{d z} = e^{-z}(z-2) + e^{-z}(1-z) = e^{-z}(z-2+1-z) = e^{-z}(-1) = -e^{-z}$.

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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