# Product rule for differentiation

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## Name

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

## Statement for two functions

### Statement in multiple versions

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:

$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = f'(x_0)g(x_0) + f(x_0)g'(x_0)$
or equivalently:
$\! \frac{d}{dx} [f(x)g(x)]|_{x = x_0} = \frac{d(f(x))}{dx}|_{x=x_0} \cdot g(x_0) + f(x_0)\cdot \frac{d(g(x))}{dx}|_{x = x_0}$

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 (see concept of equality conditional to existence of one side):
$\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
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 (see concept of equality conditional to existence of one side):
$\! (f \cdot g)' = (f'\cdot g) + (f \cdot g')$
We could also write this more briefly as:
$\! (fg)' = f'g + fg'$
Note that the domain of $(fg)'$ may be strictly larger than the intersection of the domains of $f'$ and $g'$, so the equality need not hold in the sense of equality as functions if we care about the domains of definition.
Pure Leibniz notation using dependent and independent variables Suppose $u,v$ are variables both of which are functionally dependent on $x$. Then:
$\! \frac{d(uv)}{dx} = \left(\frac{du}{dx}\right) v + u \frac{dv}{dx}$
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.

### Partial differentiation

For further information, refer: product rule for partial differentiation

The product rule is also valid if we consider functions of more than one variable and replace the ordinary derivative by the partial derivative, directional derivative, or gradient vector.

## Statement for multiple functions

Below, we formulate the many versions of this product rule:

Version type Statement
specific point, named functions Suppose $f_1, f_2, \dots, f_n$ are functions defined and differentiable at a point $x_0$. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ is also differentiable at $x_0$, and we have:
$\! \frac{d}{dx}[f_1(x)f_2(x)\dots f_n(x)]|_{x = x_0} = f_1'(x_0)f_2(x_0) \dots f_n(x_0) + f_1(x_0)f_2'(x_0) \dots f_n(x_0) + \dots + f_1(x_0)f_2(x_0) \dots f_{n-1}(x_0)f_n'(x_0)$
generic point, named functions, point notation Suppose $f_1, f_2, \dots, f_n$ are functions. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ satisfies:
$\! (f_1 \cdot f_2 \cdot \dots \cdot f_n)'(x) = f_1'(x)f_2(x) \dots f_n(x) + f_1(x)f_2'(x) \dots f_n(x) + \dots + f_1(x)f_2(x) \dots f_{n-1}(x)f_n'(x)$ wherever the right side makes sense.
generic points, named functions, point-free notation Suppose $f_1, f_2, \dots, f_n$ are functions. Then the product $f_1 \cdot f_2 \cdot \dots \cdot f_n$ satisfies:
$\! (f_1 \cdot f_2 \cdot \dots \cdot f_n)' = f_1' \cdot f_2 \cdot \dots f_n + f_1 \cdot f_2' \cdot \dots \cdot f_n + \dots + f_1 \cdot f_2 \cdot \dots \cdot f_{n-1} \cdot f_n'$ wherever the right side makes sense. We could also write this more briefly as
$\! (f_1 f_2 \dots f_n)' = f_1' f_2 \dots f_n + f_1 f_2' \dots f_n + \dots + f_1 f_2 \dots \cdot f_{n-1}f_n'$
Pure Leibniz notation using dependent and independent variables Suppose $u_1,u_2,\dots,u_n$ are variables functionally dependent on $x$. Then $\frac{d(u_1u_2\dots u_n)}{dx} = \left(\frac{du_1}{dx}\right)(u_2u_3 \dots u_n) + u_1 \left(\frac{du_2}{dx}\right) (u_3 \dots u_n) + \dots + u_1u_2 \dots u_{n-1} \left(\frac{du_n}{dx}\right)$ wherever the right side make sense.
In terms of differentials Suppose $u_1,u_2,\dots,u_n$ are variables functionally dependent on $x$. Then $d(u_1u_2\dots u_n) = u_2u_3 \dots u_n(du_1) + u_1u_3 \dots u_n(du_2) + \dots + u_1u_2 \dots u_{n-1}(du_n)$

For instance, using the generic point, named functions notation for $n = 3$, we get:

$\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)$

## Reversal for integration

The reverse to this rule, that is helpful for indefinite integrations, is a method called integration by parts.

## 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 $f \cdot g$. The one-sided versions allow us to make similar statement for left and right differentiability.
generic point, named functions, point notation This tells us that if both $f$ and $g$ are differentiable on an open interval, then so is $f \cdot g$. The one-sided versions allow us to make similar statements for closed intervals where we require the appropriate one-sided differentiability at the endpoints.
generic point, point-free notation This can be used to deduce more, namely that the nature of $(f \cdot g)'$ 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., $f'$ and $g'$ are both continuous on that interval), then $(f \cdot g)'$ 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) $\! f(x_0), g(x_0), f'(x_0), g'(x_0)$ at a specific point $x_0$ is sufficient to compute the value of $(f \cdot g)'(x_0)$. For instance, if we are given that $f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13$, we obtain that $(f \cdot g)'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109$.
A note on contrast with the (false) freshman product rule: [SHOW MORE]
generic point, named functions This tells us that knowledge of the general expressions for $f$ and $g$ and 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 $\! f'(x_0)$ and $\! g'(x_0)$ are both equal to 0, then so is $(f \cdot g)'(x_0)$. 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, but so would a stronger result: namely that if either $f'(x_0)$ or $g'(x_0)$ is zero, so is $(f \cdot g)'(x_0)$.
significance of sign of derivative $\! f'(x_0)$ and $\! g'(x_0)$ both being positive is not sufficient to ensure that $(f \cdot g)'(x_0)$ is positive. However, if all four of $\! f(x_0), g(x_0), f'(x_0), g'(x_0)$ are positive, then $(f \cdot g)'(x_0)$ 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 $\! f'(x_0)$ and $\! g'(x_0)$ both being positive is sufficient to ensure that $(f \cdot g)'(x_0)$ is positive.
significance of uniform bounds $\! f',g'$ both being uniformly bounded is not sufficient to ensure that $(f \cdot g)'$ is uniformly bounded. However, if all four functions $\! f,g,f',g'$ are uniformly bounded, then indeed $(f \cdot g)'$ is uniformly bounded. In that case, it would be true that $\! f'(x_0)$ and $\! g'(x_0)$ both uniformly being bounded is sufficient to ensure that $(f \cdot g)'(x_0)$ is uniformly bounded.

## Compatibility checks

### Symmetry in the functions being multiplied

We know that the product of two functions is symmetric in them, i.e., $f \cdot g = g \cdot f$. Thus, the product rule for differentiation should satisfy the condition that the formula for $(f \cdot g)'$ is symmetric in $f$ and $g$, i.e., we get the same formula for $(g \cdot f)'$. This is indeed true using the commutativity of addition and multiplication:

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

### Associative symmetry

This is a compatibility check showing that for a product of three functions $f_1 \cdot f_2 \cdot f_3$, we get the same product rule formula whether we associate this product as $f_1 \cdot (f_2 \cdot f_3)$ or as $(f_1 \cdot f_2) \cdot f_3$.

• Associating as $f_1 \cdot (f_2 \cdot f_3)$:

$(f_1 \cdot (f_2 \cdot f_3))' = (f_1' \cdot (f_2 \cdot f_3)) + (f_1 \cdot (f_2 \cdot f_3)') = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot (f_2' \cdot f_3 + f_2 \cdot f_3') = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'$

• Associating as $(f_1 \cdot f_2) \cdot f_3$:

$((f_1 \cdot f_2) \cdot f_3)' = (f_1 \cdot f_2)' \cdot f_3 + (f_1 \cdot f_2) \cdot f_3' = (f_1' \cdot f_2 + f_1 \cdot f_2') \cdot f_3 + f_1 \cdot f_2 \cdot f_3' = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'$

### Compatibility with linearity

Consider functions $f,g,h$ and the expression:

$\! f \cdot (g + h) = (f \cdot g) + (f \cdot h)$

This can be differentiated in two ways, using the product rule on the left side and then linearity of differentiation, or by differentiating the right side and using the product rule in each term. We verify that both yield the same result:

• Left side: Differentiating, we get $\! f' \cdot (g + h) + f \cdot (g + h)' = f' \cdot g + f' \cdot h +f \cdot g' + f \cdot h'$
• Right side: Differentiating, we get $\! (f \cdot g)' + (f \cdot h)' = f' \cdot g + f \cdot g' + f' \cdot h + f \cdot h'$.

Thus, both sides are equal and the product rule for differentiation checks out.

### Compatibility with notions of order

This section explains why the product rule is compatible with notions of order $\operatorname{ord}$ that satisfy:

• $\operatorname{ord}(f') = \operatorname{ord}(f) - 1$
• $\operatorname{ord}(f \cdot g) = \operatorname{ord}(f) + \operatorname{ord}(g)$
• If $\operatorname{ord}(f) = \operatorname{ord}(g)$, and both $f,g$ are positive (in a suitable sense) then $\operatorname{ord}(f + g)$ equals it. Even if they are not both positive, $f + g$ usually has the same order

Suppose $\operatorname{ord}(f) = m$ and $\operatorname{ord}(g) = n$. Then we have the following:

• $(f \cdot g)'$ has order $m + n -1$: First, note that $f \cdot g$ has order $m + n$ by the product relation for order. Next, note that differentiating pushes the order down by one.
• $(f' \cdot g) + (f \cdot g')$ also (plausibly) has order $m + n - 1$: Note that $f' \cdot g$ has order $(m - 1) + n = m + n -1$ and $f \cdot g' = m + (n - 1) = m + n - 1$. Adding them should give something of order $m + n - 1$.

Thus, the product rule is compatible with the order notion.

Note that the freshman product rule is incompatible with notions of order: [SHOW MORE]

Some examples of the notion of order which illustrate this are:

• For nonzero polynomials, the order notion above can be taken as the degree of the polynomial (though the zero polynomial creates some trouble for multiplication). The notion of positive can be taken as having a positive leading coefficients.
• For functions that are zero at a particular point, the order notion above can be taken as the order of zero at the point.

## Case of infinite or undefined values

The product rule for differentiation has analogues for infinities, with the appropriate caveats about indeterminate forms. Specifically, we have the following:

$f(x_0)$ $g(x_0)$ $f'(x_0)$ $g'(x_0)$ Conclusion about $(f \cdot g)'(x_0)$ Explanation
finite finite undefined undefined insufficient information (could be finite or undefined) We don't know the details behind the undefined
nonzero nonzero and same sign as $f(x_0)$ vertical tangent vertical tangent of same type as for $f$ (i.e., either both are increasing or both are decreasing) vertical tangent, type (increasing/decreasing) is determined by signs of $f,g$ and types of vertical tangent for $f,g$ [SHOW MORE]
nonzero nonzero and opposite sign to $f(x_0)$ vertical tangent vertical tangent of same type as for $f$ (i.e., either both are increasing or both are decreasing) insufficient information [SHOW MORE]
nonzero nonzero and same sign as $f(x_0)$ vertical tangent vertical tangent of opposite type as for $f$ (i.e., one is increasing and one is decreasing) insufficient information
nonzero nonzero and opposite sign to $f(x_0)$ vertical tangent vertical tangent of opposite type as for $f$ (i.e., one is increasing and one is decreasing) vertical tangent, type depends on signs
zero known whether it is zero, positive, or negative known whether it is finite, vertical tangent, etc. vertical tangent insufficient information in all cases.

## Examples

For practical tips and explanations on how to apply the product rule in practice, check out Practical:Product rule for differentiation

### Sanity checks

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

Case The derivative of $x \mapsto f(x)g(x)$ Direct justification (without use of product rule) Justification using product rule, i.e., computing it as $\! f'(x)g(x) + f(x)g'(x)$
$g$ is the zero function. zero function $\! f(x)g(x) = 0$ for all $x$, so its derivative is also zero . 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 $\lambda$. $\! \lambda f'(x)$ The function is $x \mapsto \lambda f(x)$, and 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 $\! \lambda f'(x)$.
$f = g$ $\! 2f(x)f'(x)$ The derivative is $\! 2f(x)f'(x)$ by the chain rule for differentiation: we are composing the square function and $f$. We get $\! f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x)$.
$g = 1/f$ zero function The product is $1$, which is a constant function, so its derivative is zero. We get $f(x)(1/f)'(x) + f'(x)/f(x)$. By the chain rule, $(1/f)'(x) = -f'(x)/(f(x))^2$, so plugging in, we get $-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)$, which simplifies to zero.

### Nontrivial examples where simple alternate methods exist

Here is a simple trigonometric example:

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

### 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:

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

## Proof

There are many different versions of the proof, given below: