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

Question

Aicharya · Accepted Answer

Given, $y=\frac{\log _e x}{x}$ and $z=\log _e x$ $	herefore \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 $\begin{aligned} \frac{d^2 y}{d z^2} & =\frac{e^z(-1)-(1-z) e^z}{e^{2 z}} \ & =\frac{-2+z}{e^z} \end{aligned}$ $\frac{d^2 y}{d z^2}+\frac{d y}{d z}=\frac{1-z}{e^z}+\frac{-2+z}{e^z}=-\frac{1}{e^z}=-e^{-z}$

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

Solution