Product rule for differentiation

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:

$\frac{d}{dx}[f(x)g(x)]{|}_{x=x_0}=f^{\prime }(x_0)g(x_0)+f(x_0)g^{\prime }(x_0)$

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:

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

or equivalently:

$(f\cdot g{)}^{\prime }=(f^{\prime }\cdot g)+(f\cdot g^{\prime })$

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:

$F^{\prime }(x)={f_{1^{\prime }}}^{\prime }(x)f_2(x)\dots f_n(x)+f_1(x){f_{2^{\prime }}}^{\prime }(x)\dots f_n(x)+\dots +f_1(x)f_2(x)\dots f_{n-1}(x){f_{n^{\prime }}}^{\prime }(x)$

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:

$F^{\prime }(x)={f_{1^{\prime }}}^{\prime }(x)f_2(x)f_3(x)+f_1(x){f_{2^{\prime }}}^{\prime }(x)f_3(x)+f_1(x)f_2(x){f_{3^{\prime }}}^{\prime }(x)$

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

Nontrivial examples where simple alternate methods exist

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Nontrivial examples where simple alternate methods do not exist

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