Product rule for differentiation

Name

This statement is called the product rule, product rule for differentiation, or Leibniz rule.

Statement for two functions

Verbal statement

If two (possibly equal) functions are differentiable at a given real number, then their pointwise product is also differentiable at that number and the derivative of the product is the sum of two terms: the derivative of the first function times the second function and the first function times the derivative of the second function.

Statement with symbols

Suppose $f$ and $g$ are functions, both of which are differentiable at a real number $x=x_{0}$ . Then, the product function $f\cdot g$ , defined as $x\mapsto f(x)g(x)$ is also differentiable at $x$ , and the derivative at $x_{0}$ is given as follows:

$\!{\frac {d}{dx}}[f(x)g(x)]|_{x=x_{0}}=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})$

or equivalently:

$\!{\frac {d}{dx}}[f(x)g(x)]|_{x=x_{0}}={\frac {d(f(x))}{dx}}|_{x=x_{0}}\cdot g(x_{0})+f(x_{0})\cdot {\frac {d(g(x))}{dx}}|_{x=x_{0}}$

If we consider the general expressions rather than evaluation at a particular point $x_{0}$ , we can rewrite the above as:

$\!{\frac {d}{dx}}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$

or equivalently:

$(f\cdot g)'=(f'\cdot g)+(f\cdot g')$

Statement for multiple functions

If $f_{1},f_{2},\dots ,f_{n}$ are all functions, and we define $F(x):=f_{1}(x)f_{2}(x)\dots f_{n}(x)$ , then we have:

$F'(x)=f_{1}'(x)f_{2}(x)\dots f_{n}(x)+f_{1}(x)f_{2}'(x)\dots f_{n}(x)+\dots +f_{1}(x)f_{2}(x)\dots f_{n-1}(x)f_{n}'(x)$

In other words, we get a sum of $n$ terms, each of which is a product of $n$ evaluations, of which only one is a derivative, and the one we choose as the derivative cycles through all the $n$ possibilities.

For instance, if $n=3$ , we get:

$\!F'(x)=f_{1}'(x)f_{2}(x)f_{3}(x)+f_{1}(x)f_{2}'(x)f_{3}(x)+f_{1}(x)f_{2}(x)f_{3}'(x)$

Related rules

Differentiation is linear: The derivative of the sum is the sum of the derivatives, and scalars can be pulled out of differentiation.
Chain rule for differentiation
Product rule for higher derivatives
Chain rule for higher derivatives

Examples

Trivial examples

We first consider examples where the product rule for differentiation confirms something we already knew through other means:

Case	What we know about the derivative of $x\mapsto f(x)g(x)$	What we know about $f'(x)g(x)+f(x)g'(x)$
$g$ is the zero function.	The derivative is the zero function, because $f(x)g(x)=0$ for all $x$ .	Both $g(x)$ and $g'(x)$ are zero functions, so $f'(x)g(x)+f(x)g'(x)$ is everywhere zero.
$g$ is a constant nonzero function with value $\lambda$ .	The derivative is $\lambda f'(x)$ , because the constant can be pulled out of the differentiation process.	$f'(x)g(x)$ simplifies to $\lambda f'(x)$ . Since $g$ is constant, $g'(x)$ is the zero function, hence so is $f(x)g'(x)$ . The sum is thus $\lambda f'(x)$ .
$f=g$	The derivative is $2f(x)f'(x)$ by the chain rule for differentiation: we are composing the square function and $f$ .	We get $f'(x)f(x)+f(x)f'(x)=2f(x)f'(x)$ .