Increase, decrease and graph shape (exp and log functions)

Differentiating $y=x^n e^{-x}$ with the product rule lets you factor out $e^{-x}$.

Since $e^{-x}>0$, the sign of $y'$ is decided by $x^{n-1}(n-x)$ alone. For $y=xe^{-x}$, $y'=(1-x)e^{-x}$ changes from positive to negative at $x=1$, so there is a local maximum $\frac{1}{e}$ at $x=1$.

For a function with $\log x$, first check the domain $x>0$. For $y=\frac{\log x}{x}$, $y'=\frac{1-\log x}{x^2}$, and the local maximum is $\frac{1}{e}$ at $x=e$.

Examples

1. 01For which x is y'=0? (other than 0)$\displaystyle y=x^{2}e^{-x},\; y'=x^{1}(2-x)e^{-x}$AnswerHide

   Answer$\displaystyle 2$
2. 02Differentiate:$\displaystyle y=x^{3}e^{-x}$AnswerHide

   Answer$\displaystyle (3x^{2}-x^{3})e^{-x}$
3. 03What is the domain?$\displaystyle y=x\log_e x$AnswerHide

   Answer$\displaystyle x > 0$
4. 04For which x is y'=0?$\displaystyle y=xe^{-x},\; y'=(1-x)e^{-x}$AnswerHide

   Answer$\displaystyle 1$
5. 05Differentiate:$\displaystyle y=x^{2}e^{-x}$AnswerHide

   Answer$\displaystyle (2x-x^{2})e^{-x}$
6. 06Differentiate:$\displaystyle y=\frac{\log_e x}{x}$AnswerHide

   Answer$\displaystyle \frac{1-\log_e x}{x^{2}}$

FAQ

Q1How do I find the sign of $y'=(1-x)e^{-x}$?AnswerHide

A

$e^{-x}$ is always positive, so the sign depends only on $1-x$: positive for $x<1$ and negative for $x>1$.

Q2Does the sign table for $y=x\log x$ include $x<0$?AnswerHide

A

No. $\log x$ is defined only for $x>0$, so the table covers $x>0$ only.