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 $z$: $\frac{d y}{d z} = \frac{d}{d z}(z e^{-z}) = e^{-z} + z(-e^{-z}) = e^{-z}(1-z)$. 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

Similar Questions

If $y=3 e^{5 x}+5 e^{3 x}$,then $\frac{d^{2} y}{d x^{2}}-8 \frac{d y}{d x}=$ (in $y$)

If $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$,then find the $n^{th}$ derivative $y_n$.

$A$ polynomial $P(x)$ with real coefficients has the property that $P^{\prime \prime}(x) \neq 0$ for all $x$. Suppose $P(0) = 1$ and $P^{\prime}(0) = -1$. What can you say about $P(1)$?

If $f(x)=x^4-x^3+7x^2+14$,then what is the value of $f^{\prime \prime}(5)$?

If $y=a \cos (\sin 2 x)+b \sin (\sin 2 x)$,then $y^{\prime \prime}+(2 \tan 2 x) y^{\prime}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module