The product rule

A function in the form of a product is not differentiated by differentiating each factor and multiplying¹. We check it with $f(x) = xe^{-x}$ .

The formula

(uv)' = u'v + uv'

Taking $u = x$ and $v = e^{-x}$ gives $u' = 1$ and $v' = -e^{-x}$ , so the derivative is as follows.

f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = (1 - x)e^{-x}

Note that this is not the same as $u'v' = -e^{-x}$ .

Monotonicity

Since $e^{-x} > 0$ , the sign is decided by $1 - x$ alone.

Range	$f'$	Behavior
$x < 1$	positive	increasing
$x > 1$	negative	decreasing

At $x = 1$ the function has its local maximum $f(1) = \dfrac{1}{e} \approx 0.368$ .

Behavior far out

As $x \to \infty$ the decay of $e^{-x}$ beats the growth of $x$ , so $f \to 0$ . As $x \to -\infty$ the factor $x$ is negative and $e^{-x}$ is large, so $f \to -\infty$ .

Differentiating once more

The same formula gives the second derivative.

f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}

The sign changes at $x = 2$ , so the inflection point is $\left( 2, \dfrac{2}{e^2} \right)$ , numerically $(2, 0.271)$ .

$x$	$f(x)$	Position
$0$	$0$	passes through the origin
$1$	$\dfrac{1}{e} \approx 0.368$	local maximum
$2$	$\dfrac{2}{e^2} \approx 0.271$	inflection point

Why the formula holds

The area of a rectangle makes it visible. For a rectangle with sides $u$ and $v$ , increasing $u$ by $\Delta u$ and $v$ by $\Delta v$ increases the area by the following.

u\Delta v + v\Delta u + \Delta u \Delta v

The last term is a product of two small quantities, so dividing by $\Delta x$ and letting it go to $0$ makes it vanish. What remains is $u'v + uv'$ .

Applications

Functions of this form appear wherever a quantity grows while decaying. The expression $xe^{-x}$ is used as a distribution of waiting times, and the position $x = 1$ of its peak is the most likely value.

The humped curve on the graph is $y = xe^{-x}$ , the curve crossing it is the derivative, and the large dots are the local maximum and the origin.

Product rule, Wikipedia