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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $y=\frac{z}{e^z} = z e^{-z}$.

Vedclass · Accepted Answer

$-e^{-z}$

Vedclass · Answer

(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

Similar Questions

If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{235} \text{U}$,then the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

$DNA$ fingerprinting method is very useful for:

Copper sulphate is prepared by blowing a current of air through copper scrap and dilute $H_2SO_4$. Dilute $HNO_3$ is also added:

Vedclass Test Series

Exam Paper Generator

Online Exam Module