If y=frac{log _e x}{x} and z=log _e x,then fr | 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 $z$: $\frac{d y}{d z} = \frac{d}{d z}(z e^{-z}) = e^{-z} - z e^{-z} = (1-z)e^{-z}$. Next,differentiating $\frac{d y}{d z}$ with respect to $z$ to find the second derivative: $\frac{d^2 y}{d z^2} = \frac{d}{d z}((1-z)e^{-z}) = -e^{-z} - (1-z)e^{-z} = (-1-1+z)e^{-z} = (z-2)e^{-z}$. Finally,calculating the sum: $\frac{d^2 y}{d z^2} + \frac{d y}{d z} = (z-2)e^{-z} + (1-z)e^{-z} = (z-2+1-z)e^{-z} = -1 \cdot e^{-z} = -e^{-z}$.

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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