# Thread: Differentiate Exponents

1. ## Differentiate Exponents

Find first and second derivatives, critical number, inflection points.

f(x) = xe^x
f'(x) = ('x) (e^x) + x(e^x)('x)
f'(x) = e^x+xe^x
f'(x) = e^x(1+x)
Critical number x = -1

f"(x) = e^x(1) + (1+x)(e^x)
f"(x) = e^x +e^x(1+x)
Critical Number x=-1

I hope I am close to the correct answer.

2. Originally Posted by confusedagain
f"(x) = e^x(1) + (1+x)(e^x)
f"(x) = e^x +e^x(1+x)
Critical Number x=-1

I hope I am close to the correct answer.
first of all, we do not find critical points from the second derivative, we use the second derivative to find concavity and possible inflection points. second of all,
$\displaystyle f''(-1) = \frac {1}{e} \neq 0$

3. Hello, confusedagain!

A small error . . .

Find first and second derivatives, critical number, inflection points.
. . $\displaystyle f(x) \:= \:xe^x$

Your first part is correct:

. . $\displaystyle f'(x) \:= \:1\cdot e^x + x\cdot e^x \:=\:e^x(1 + e^x)$

Then: .$\displaystyle e^x(1+x) \:=\:0\quad\Rightarrow\quad x = -1$

. . Critical point: $\displaystyle \left(-1,\:-\frac{1}{e}\right)$

$\displaystyle f''(x) \:= \:e^x + (1+x)e^x \:=\:e^x(1 + 1 + x) \:=\:e^x(2 + x)$

Then: .$\displaystyle e^x(2 + x) \:=\:0\quad\Rightarrow\quad x = -2$

. . Inflection point: .$\displaystyle \left(-2,\:-\frac{2}{e^2}\right)$

4. Hey Soroban

e^x(2 + x) \:=\:0\quad\Rightarrow\quad x = -2

If you prefer, use e^x(2 + x)=0\implies x=-2, this yields $\displaystyle e^x(2 + x)=0\implies x=-2$