# Product rule

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In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated as

$(f\cdot g)'=f'\cdot g+f\cdot g'\,\!$ or in the Leibniz notation

${\dfrac {d}{dx}}(u\cdot v)=u\cdot {\dfrac {dv}{dx}}+v\cdot {\dfrac {du}{dx}}$ .

In the notation of differentials this can be written as

$d(uv)=u\,dv+v\,du$ .

In Leibniz notation, the derivative of the product of three functions (not to be confused with Euler's triple product rule) is

${\dfrac {d}{dx}}(u\cdot v\cdot w)={\dfrac {du}{dx}}\cdot v\cdot w+u\cdot {\dfrac {dv}{dx}}\cdot w+u\cdot v\cdot {\dfrac {dw}{dx}}$ .

## Discovery

Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials. (However, Child (2008) argues that it is due to Isaac Barrow). Here is Leibniz's argument: Let u(x) and v(x) be two differentiable functions of x. Then the differential of uv is

{\begin{aligned}d(u\cdot v)&{}=(u+du)\cdot (v+dv)-u\cdot v\\&{}=u\cdot dv+v\cdot du+du\cdot dv.\end{aligned}} Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that

$d(u\cdot v)=v\cdot du+u\cdot dv\,\!$ and this is indeed the differential form of the product rule. If we divide through by the differential dx, we obtain

${\frac {d}{dx}}(u\cdot v)=v\cdot {\frac {du}{dx}}+u\cdot {\frac {dv}{dx}}\,\!$ which can also be written in Lagrange's notation as

$(u\cdot v)'=v\cdot u'+u\cdot v'.\,\!$ ## Examples

• Suppose we want to differentiate ƒ(x) = x2 sin(x). By using the product rule, one gets the derivative ƒ '(x) = 2x sin(x) + x2cos(x) (since the derivative of x2 is 2x and the derivative of sin(x) is cos(x)).
• One special case of the product rule is the constant multiple rule which states: if c is a real number and ƒ(x) is a differentiable function, then (x) is also differentiable, and its derivative is (c × ƒ)'(x) = c × ƒ '(x). This follows from the product rule since the derivative of any constant is zero. This, combined with the sum rule for derivatives, shows that differentiation is linear.
• The rule for integration by parts is derived from the product rule, as is (a weak version of) the quotient rule. (It is a "weak" version in that it does not prove that the quotient is differentiable, but only says what its derivative is if it is differentiable.)

## Proofs

### Simple Proof

Let h(x) = f(x) g(x), and suppose that f and g are each differentiable at x. We want to prove that h is differentiable at x and that its derivative h'(x) is given by f'(x) g(x) + f(x) g'(x).

$=\lim _{a\to 0}{\frac {f(x+a)g(x+a)-f(x)g(x+a)+f(x)g(x+a)-f(x)g(x)}{a}}$ $=\lim _{a\to 0}{\frac {[f(x+a)-f(x)]\cdot g(x+a)+f(x)\cdot [g(x+a)-g(x)]}{a}}$ $=\lim _{a\to 0}{\frac {f(x+a)-f(x)}{a}}\cdot \lim _{a\to 0}g(x+a)+\lim _{a\to 0}f(x)\cdot \lim _{a\to 0}{\frac {g(x+a)-g(x)}{a}}$ $=f'(x)g(x)+f(x)g'(x)$ .

### More Complicated Proof

{{ safesubst:#invoke:Unsubst||$N=Unreferenced section |date=__DATE__ |$B= {{ safesubst:#invoke:Unsubst||$N=Unreferenced |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} }} A rigorous proof of the product rule can be given using the definition of the derivative as a limit, and the basic properties of limits.

Let h(x) = f(x) g(x), and suppose that f and g are each differentiable at x0. (Note that x0 will remain fixed throughout the proof). We want to prove that h is differentiable at x0 and that its derivative h'(x0) is given by f'(x0) g(x0) + f(x0) g'(x0).

Let Δh = h(x0+Δx) - h(x0); note that although x0 is fixed, Δh depends on the value of Δx, which is thought of as being "small."

The function h is differentiable at x0 if the limit

$\lim _{\Delta x\to 0}{\Delta h \over \Delta x}$ exists; when it does, h'(x0) is defined to be the value of the limit.

As with Δh, let Δf = f(x0+Δx) - f(x0) and Δg = g(x0+Δx) - g(x0) which, like Δh, also depends on Δx. Then f(x0+Δx) = f(x0) + Δf and g(x0+Δx) = g(x0) + Δg.

It follows that h(x0+Δx) = f(x0+Δx) g(x0+Δx) = (f(x0) + Δf) (g(x0)+Δg); applying the distributive law, we see that

While it is not necessary for the proof, it can be helpful to understand this product geometrically as the area of the rectangle in this diagram:

To get the value of Δh, subtract h(x0)=f(x0) g(x0) from equation Template:EquationNote. This removes the area of the white rectangle, leaving three rectangles:

$\Delta h=\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g$ To find h'(x0), we need to find the limit as Δx goes to 0 of

The first two terms of the right-hand side of this equation correspond to the areas of the blue rectangles; the third corresponds to the area of the gray rectangle. Using the basic properties of limits and the definition of the derivative, we can tackle this term-by term. First,

$\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}g(x_{0})\right)=f'(x_{0})g(x_{0})$ .

Similarly,

$\lim _{\Delta x\to 0}\left(f(x_{0}){\frac {\Delta g}{\Delta x}}\right)=f(x_{0})g'(x_{0})$ .

The third term, corresponding to the small gray rectangle, winds up being negligible (i.e. going to 0 in the limit) because Δf Δg "vanishes to second order." Rigorously,

$\lim _{\Delta x\to 0}{\frac {\Delta f\Delta g}{\Delta x}}=\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}{\frac {\Delta g}{\Delta x}}\Delta x\right)=\lim _{\Delta x\to 0}{\frac {\Delta f}{\Delta x}}\cdot \lim _{\Delta x\to 0}{\frac {\Delta g}{\Delta x}}\cdot \lim _{\Delta x\to 0}{\Delta x}=f'(x_{0})g'(x_{0})\cdot 0=0$ We have shown that the limit of each of the three terms on the right-hand side of equation Template:EquationNote exists, hence

$\lim _{\Delta x\to 0}{\frac {\Delta h}{\Delta x}}$ exists and is equal to the sum of the three limits. Thus, the product h(x) is differentiable at x0 and its derivative is given by

{\begin{aligned}h'(x_{0})&=\lim _{\Delta x\to 0}{\frac {\Delta h}{\Delta x}}\\&=\lim _{\Delta x\to 0}\left({\frac {\Delta f}{\Delta x}}g(x_{0})\right)+\lim _{\Delta x\to 0}\left(f(x_{0}){\frac {\Delta g}{\Delta x}}\right)+\lim _{\Delta x\to 0}\left({\frac {\Delta f\Delta g}{\Delta x}}\right)\\&=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})+0\\&=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})\\\end{aligned}} as was to be shown.

### Brief proof

$f(x+h)=f(x)+f'(x)h+\psi _{1}(h)\qquad \qquad g(x+h)=g(x)+g'(x)h+\psi _{2}(h)$ {\begin{aligned}fg(x+h)-fg(x)=(f(x)+f'(x)h+\psi _{1}(h))(g(x)+g'(x)h+\psi _{2}(h))-fg(x)=f'(x)g(x)h+f(x)g'(x)h+O(h)\\[12pt]\end{aligned}} Taking the limit for small $h$ gives the result.

### Logarithms and quarter squares

Let f = uv and suppose u and v are positive functions of x. Then

$\ln f=\ln(u\cdot v)=\ln u+\ln v.\,$ Differentiating both sides:

${1 \over f}{df \over dx}={1 \over u}{du \over dx}+{1 \over v}{dv \over dx}\,$ and so, multiplying the left side by f, and the right side by uv (note: f = uv),

${df \over dx}=v{du \over dx}+u{dv \over dx}.\,$ The proof appears in . Note that since u, v need to be continuous, the assumption on positivity does not diminish the generality.

This proof relies on the chain rule and on the properties of the natural logarithm function, both of which are deeper than the product rule (however, information about the derivative of a logarithm that is sufficient to carry out a variant of the proof can be inferred by considering the derivative at x = 1 of the logarithm to any base of cx, where c is a constant, then generalising c). From one point of view, that is a disadvantage of this proof. On the other hand, the simplicity of the algebra in this proof perhaps makes it easier to understand than a proof using the definition of differentiation directly.

There is an analogous but arguably even easier proof (i.e., some people may find it easier as it can be used before being able to differentiate logarithms), using quarter square multiplication, which similarly relies on the chain rule and on the properties of the quarter square function (shown here as q, i.e., with $q(x)={x^{2} \over 4}$ ):

$f=q(u+v)-q(u-v),$ Differentiating both sides:

$f'=q'(u+v)(u'+v')-q'(u-v)(u'-v')$ $=\left({1 \over 2}(u+v)(u'+v')\right)-\left({1 \over 2}(u-v)(u'-v')\right)$ $={1 \over 2}(uu'+vu'+uv'+vv')-{1 \over 2}(uu'-vu'-uv'+vv')$ $=vu'+uv'$ $=uv'+u'v.\,$ This does not present issues of whether the values are positive or negative, and the function's properties are much simpler to demonstrate (indeed, it can be differentiated without using first principles by considering the derivative at x = 0 of cx, where c is a constant, then generalising c).

Note also, these proofs are only valid for numbers or similar, whereas proofs from first principles are also valid for matrices and such like.

### Chain rule

The product rule can be considered a special case of the chain rule for several variables.

${d(ab) \over dx}={\frac {\partial (ab)}{\partial a}}{\frac {da}{dx}}+{\frac {\partial (ab)}{\partial b}}{\frac {db}{dx}}=b{\frac {da}{dx}}+a{\frac {db}{dx}}.\,$ ### Non-standard analysis

Let u and v be continuous functions in x, and let dx, du and dv be infinitesimals within the framework of non-standard analysis, specifically the hyperreal numbers. Using st to denote the standard part function that associates to a finite hyperreal number the real infinitely close to it, this gives

This was essentially Leibniz's proof exploiting the transcendental law of homogeneity (in place of the standard part above).

### Smooth infinitesimal analysis

In the context of Lawvere's approach to infinitesimals, let dx be a nilsquare infinitesimal. Then du = u' dx and dv = v' dx, so that

{\begin{aligned}d(uv)&{}=(u+du)(v+dv)-uv\\&{}=uv+u\cdot dv+v\cdot du+du\cdot dv-uv\\&{}=u\cdot dv+v\cdot du+du\cdot dv\\&{}=u\cdot dv+v\cdot du\,\!\end{aligned}} since

$du\cdot dv=u'v'(dx)^{2}=0\,\!$ ## Generalizations

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### A product of more than two factors

The product rule can be generalized to products of more than two factors. For example, for three factors we have

${\frac {d(uvw)}{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}\,\!$ .
${\frac {d}{dx}}\left[\prod _{i=1}^{k}f_{i}(x)\right]=\sum _{i=1}^{k}\left({\frac {d}{dx}}f_{i}(x)\prod _{j\neq i}f_{j}(x)\right)=\left(\prod _{i=1}^{k}f_{i}(x)\right)\left(\sum _{i=1}^{k}{\frac {f'_{i}(x)}{f_{i}(x)}}\right).$ ### Higher derivatives

It can also be generalized to the Leibniz rule for the nth derivative of a product of two factors:

$(uv)^{(n)}(x)=\sum _{k=0}^{n}{n \choose k}\cdot u^{(n-k)}(x)\cdot v^{(k)}(x).$ Furthermore, for the nth derivative of an arbitrary number of factors:

$\left(\prod _{i=1}^{k}f_{i}\right)^{(n)}=\sum _{j_{1}+j_{2}+...+j_{k}=n}{n \choose j_{1},j_{2},...,j_{k}}\prod _{i=1}^{k}f_{i}^{(j_{i})}.$ ### Higher partial derivatives

For partial derivatives, we have

${\partial ^{n} \over \partial x_{1}\,\cdots \,\partial x_{n}}(uv)=\sum _{S}{\partial ^{|S|}u \over \prod _{i\in S}\partial x_{i}}\cdot {\partial ^{n-|S|}v \over \prod _{i\not \in S}\partial x_{i}}$ where the index S runs through the whole list of 2n subsets of {1, ..., n}. For example, when n = 3, then

$(D_{\left(x,y\right)}\,B)\left(u,v\right)=B\left(u,y\right)+B\left(x,v\right)\qquad \forall (u,v)\in X\times Y.$