Product rule for differentiation

This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules

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

The product rule is stated in many versions:

Version type	Statement
specific point, named functions	Suppose $f$ and $g$ are functions of one variable, 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 $x=x_0$, 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: $\frac{d}{dx}[f(x)g(x)]{|}_{x=x_0}=\frac{d(f(x))}{dx}{|}_{x=x_0}\cdot g(x_0)+f(x_0)\cdot \frac{d(g(x))}{dx}{|}_{x=x_0}$
generic point, named functions, point notation	Suppose $f$ and $g$ are functions of one variable. Then the following is true wherever the right side expression makes sense: $\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$
generic point, named functions, point-free notation	Suppose $f$ and $g$ are functions of one variable. Then, we have the following equality of functions on the domain where the right side expression makes sense: $(f\cdot g{)}^{\prime }=(f^{\prime }\cdot g)+(f\cdot g^{\prime })$
Pure Leibniz notation using dependent and independent variables	Suppose $u,v$ are variables both of which are functionally dependent on $x$. Then: $\frac{d(uv)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$
In terms of differentials	Suppose $u,v$ are both variables functionally dependent on $x$. Then, $d(uv)=v(du)+u(dv)$.

MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a

$\{{\}}_0$

subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.

One-sided version

The product rule for differentiation has analogues for one-sided derivatives. More explicitly, we can replace all occurrences of derivatives with left hand derivatives and the statements are true. Alternately, we can replace all occurrences of derivatives with right hand derivatives and the statements are true.

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

Significance

Qualitative and existential significance

Each of the versions has its own qualitative significance:

Version type	Significance
specific point, named functions	This tells us that if $f$ and $g$ are both differentiable at a point, so is $f\cdot g$.
generic point, named functions, point notation	This tells us that if both $f$ and $g$ are differentiable on an open interval, then so is $f\cdot g$.
generic point, point-free notation	This can be used to deduce more, namely that the nature of $(f\cdot g{)}^{\prime }$ depends strongly on the nature of $f$ and that of $g$. In particular, if $f$ and $g$ are both continuously differentiable functions on an interval (i.e., $f^{\prime }$ and $g^{\prime }$ are both continuous on that interval), then $(f\cdot g{)}^{\prime }$ is also continuously differentiable on that interval. This uses the sum theorem for continuity and product theorem for continuity.

Computational feasibility significance

Each of the versions has its own computational feasibility significance:

Version type	Significance
specific point, named functions	This tells us that knowledge of the values (in the sense of numerical values) $f(x_0),g(x_0),f^{\prime }(x_0),g^{\prime }(x_0)$ at a specific point $x_0$ is sufficient to compute the value of $(f\cdot g{)}^{\prime }(x_0)$. For instance, if we are given that $f(1)=5,g(1)=11,f^{\prime }(1)=4,g^{\prime }(1)=13$, we obtain that $(f\cdot g{)}^{\prime }(1)=4\cdot 11+5\cdot 13=44+65=109$. A note on contrast with the (false) freshman product rule: [SHOW MORE] If the freshman product rule were true, then knowledge of $f^{\prime }(x_0)$ and $g^{\prime }(x_0)$ alone (without knowledge of $f(x_0)$ and $g(x_0)$) would be sufficient to compute $(f\cdot g{)}^{\prime }(x_0)$. This is not the case.
generic point, named functions	This tells us that knowledge of the general expressions for the derivatives of $f$ and $g$ is sufficient to compute the general expression for the derivative of $f\cdot g$. See the #Examples section of this page for more examples.

Computational results significance

Each of the versions has its own computational results significance:

Shorthand	Significance	What would happen if the freshman product rule were true instead of the product rule?
significance of derivative being zero	If $f^{\prime }(x_0)$ and $g^{\prime }(x_0)$ are both equal to 0, then so is $(f\cdot g{)}^{\prime }(x_0)$. In other words, if the tangents to the graphs of $f,g$ are both horizontal at the point $x=x_0$, so is the tangent to the graph of $f\cdot g$.	This result would still hold
significance of sign of derivative	$f^{\prime }(x_0)$ and $g^{\prime }(x_0)$ both being positive is not sufficient to ensure that $(f\cdot g{)}^{\prime }(x_0)$ is positive. However, if all four of $f(x_0),g(x_0),f^{\prime }(x_0),g^{\prime }(x_0)$ are positive, then $(f\cdot g{)}^{\prime }(x_0)$ is positive. This is related to the fact that a product of increasing functions need not be increasing.	In that case, it would be true that $f^{\prime }(x_0)$ and $g^{\prime }(x_0)$ both being positive is sufficient to ensure that $(f\cdot g{)}^{\prime }(x_0)$ is positive.
significance of uniform bounds	$f^{\prime },g^{\prime }$ both being uniformly bounded is not sufficient to ensure that $(f\cdot g{)}^{\prime }$ is uniformly bounded. However, if all four functions $f,g,f^{\prime },g^{\prime }$ are uniformly bounded, then indeed $(f\cdot g{)}^{\prime }$ is uniformly bounded.	In that case, it would be true that $f^{\prime }(x_0)$ and $g^{\prime }(x_0)$ both uniformly being bounded is sufficient to ensure that $(f\cdot g{)}^{\prime }(x_0)$ is uniformly bounded.

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

Here is a simple trigonometric example:

$f(x):=\sin x\cos x$.

[SHOW MORE]

Doing this using the product rule, we get:

$f^{\prime }(x)={{\sin }^{\prime }}^{\prime }x\cos x+\sin x{{\cos }^{\prime }}^{\prime }x=\cos x\cos x+\sin x(-\sin x)={\cos }^{2}x-{\sin }^{2}x=\cos (2x)$

where the last step uses a trigonometric identity.

We could also do this by a different method, noting that, by a trigonometric identity:

$f(x)=\frac{1}{2}\sin (2x)$

and now differentiate, to get:

$f^{\prime }(x)=\frac{1}{2}\frac{d}{dx}\sin (2x)=\frac{1}{2}\cos (2x)\frac{d}{dx}(2x)=\frac{1}{2}\cos (2x)(2)=\cos (2x)$

Nontrivial examples where simple alternate methods do not exist

Consider a product of the form:

$f(x):=x\sin x$

Using the product rule, we get:

$f^{\prime }(x)=\frac{dx}{dx}\sin x+x\frac{d}{dx}(\sin x)=1(\sin x)+x\cos x=\sin x+x\cos x$