Product rule

For Euler's chain rule relating partial derivatives of three independent variables, see Triple product rule. For the counting principle in combinatorics, see Rule of product.

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

In differentials notation, this can be written as

$d(uv)=u\,dv+v\,du$$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}}$${\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}}$${\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\,\!$$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}}\,\!$${\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'.\,\!$$(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 cƒ(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

Proof by factoring

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). To do this $f(x)g(x+\Delta {x})-f(x)g(x+\Delta {x})$$f(x)g(x+\Delta {x})-f(x)g(x+\Delta {x})$ (which is zero, and thus does not change the value) is added to the numerator to permit its factoring, and then properties of limits are used.

$h'(x)=\lim _{\Delta {x}\to 0}{\frac {h(x+\Delta {x})-h(x)}{\Delta {x}}}=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}$$h'(x)=\lim _{\Delta {x}\to 0}{\frac {h(x+\Delta {x})-h(x)}{\Delta {x}}}=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}$

$=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x+\Delta {x})+f(x)g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}$$=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x+\Delta {x})+f(x)g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}$

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

$=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})-f(x)}{\Delta {x}}}\cdot \lim _{\Delta {x}\to 0}g(x+\Delta {x})+\lim _{\Delta {x}\to 0}f(x)\cdot \lim _{\Delta {x}\to 0}{\frac {g(x+\Delta {x})-g(x)}{\Delta {x}}}$$=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})-f(x)}{\Delta {x}}}\cdot \lim _{\Delta {x}\to 0}g(x+\Delta {x})+\lim _{\Delta {x}\to 0}f(x)\cdot \lim _{\Delta {x}\to 0}{\frac {g(x+\Delta {x})-g(x)}{\Delta {x}}}$

$=f'(x)g(x)+f(x)g'(x)$$=f'(x)g(x)+f(x)g'(x)$.

Rigorous proof

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}$$\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

$h(x_{0}+\Delta x)=f(x_{0}+\Delta x)g(x_{0}+\Delta x)=f(x_{0})g(x_{0})+\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g$$h(x_{0}+\Delta x)=f(x_{0}+\Delta x)g(x_{0}+\Delta x)=f(x_{0})g(x_{0})+\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g$

( )

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 (*). 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$$\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

${\frac {\Delta h}{\Delta x}}={\frac {\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g}{\Delta x}}={\frac {\Delta f}{\Delta x}}g(x_{0})+f(x_{0}){\frac {\Delta g}{\Delta x}}+{\frac {\Delta f\Delta g}{\Delta x}}$${\frac {\Delta h}{\Delta x}}={\frac {\Delta fg(x_{0})+f(x_{0})\Delta g+\Delta f\Delta g}{\Delta x}}={\frac {\Delta f}{\Delta x}}g(x_{0})+f(x_{0}){\frac {\Delta g}{\Delta x}}+{\frac {\Delta f\Delta g}{\Delta x}}$

( )

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}).$$\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}).$$\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$$\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 (**) exists, hence

$\lim _{\Delta x\to 0}{\frac {\Delta h}{\Delta x}}$$\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}}$${\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

By definition, if $f,g:\mathbb {R} \rightarrow \mathbb {R}$$f,g:\mathbb {R} \rightarrow \mathbb {R}$ are differentiable at $x$$x$ then we can write

$f(x+h)=f(x)+f'(x)h+\psi _{1}(h)\qquad \qquad g(x+h)=g(x)+g'(x)h+\psi _{2}(h)$$f(x+h)=f(x)+f'(x)h+\psi _{1}(h)\qquad \qquad g(x+h)=g(x)+g'(x)h+\psi _{2}(h)$

such that $\lim _{h\to 0}{\frac {\psi _{1}(h)}{h}}=\lim _{h\to 0}{\frac {\psi _{2}(h)}{h}}=0$$\lim _{h\to 0}{\frac {\psi _{1}(h)}{h}}=\lim _{h\to 0}{\frac {\psi _{2}(h)}{h}}=0$, also written $\psi _{1},\psi _{2}\sim o(h)$$\psi _{1},\psi _{2}\sim o(h)$. Then:

${\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}}$${\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$$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.\,$$\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}\,$${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}.\,$${df \over dx}=v{du \over dx}+u{dv \over dx}.\,$

The proof appears in [1]. 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}$$q(x)={x^{2} \over 4}$):

$f=q(u+v)-q(u-v),$$f=q(u+v)-q(u-v),$

Differentiating both sides:

$f'=q'(u+v)(u'+v')-q'(u-v)(u'-v')$$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)$$=\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')$$={1 \over 2}(uu'+vu'+uv'+vv')-{1 \over 2}(uu'-vu'-uv'+vv')$

$=vu'+uv'$$=vu'+uv'$

$=uv'+u'v.\,$$=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}}.\,$${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

${\frac {d(uv)}{dx}}\,$${\frac {d(uv)}{dx}}\,$	$=\operatorname {st} \left({\frac {(u+\mathrm {d} u)(v+\mathrm {d} v)-uv}{\mathrm {d} x}}\right)$$=\operatorname {st} \left({\frac {(u+\mathrm {d} u)(v+\mathrm {d} v)-uv}{\mathrm {d} x}}\right)$
	$=\operatorname {st} \left({\frac {uv+u\cdot \mathrm {d} v+v\cdot \mathrm {d} u+\mathrm {d} v\cdot \mathrm {d} u-uv}{\mathrm {d} x}}\right)$$=\operatorname {st} \left({\frac {uv+u\cdot \mathrm {d} v+v\cdot \mathrm {d} u+\mathrm {d} v\cdot \mathrm {d} u-uv}{\mathrm {d} x}}\right)$
	$=\operatorname {st} \left({\frac {u\cdot \mathrm {d} v+(v+\mathrm {d} v)\cdot \mathrm {d} u}{\mathrm {d} x}}\right)$$=\operatorname {st} \left({\frac {u\cdot \mathrm {d} v+(v+\mathrm {d} v)\cdot \mathrm {d} u}{\mathrm {d} x}}\right)$
	$={u}{\frac {dv}{dx}}+{v}{\frac {du}{dx}}$$={u}{\frac {dv}{dx}}+{v}{\frac {du}{dx}}$.

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

Generalizations

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(uvw)}{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}\,\!$.

For a collection of functions $f_{1},\dots ,f_{k}$$f_{1},\dots ,f_{k}$, we have

${\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).$${\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).$$(uv)^{(n)}(x)=\sum _{k=0}^{n}{n \choose k}\cdot u^{(n-k)}(x)\cdot v^{(k)}(x).$

See also binomial coefficient and the formally quite similar binomial theorem. See also General Leibniz rule.

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})}.$$\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}}$${\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

${\begin{aligned}&{}\quad {\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}(uv)\\\\&{}=u\cdot {\partial ^{3}v \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{1}}\cdot {\partial ^{2}v \over \partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{2}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{3}}+{\partial u \over \partial x_{3}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{2}}\\\\&{}\qquad +{\partial ^{2}u \over \partial x_{1}\,\partial x_{2}}\cdot {\partial v \over \partial x_{3}}+{\partial ^{2}u \over \partial x_{1}\,\partial x_{3}}\cdot {\partial v \over \partial x_{2}}+{\partial ^{2}u \over \partial x_{2}\,\partial x_{3}}\cdot {\partial v \over \partial x_{1}}+{\partial ^{3}u \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}\cdot v.\end{aligned}}$${\begin{aligned}&{}\quad {\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}(uv)\\\\&{}=u\cdot {\partial ^{3}v \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{1}}\cdot {\partial ^{2}v \over \partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{2}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{3}}+{\partial u \over \partial x_{3}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{2}}\\\\&{}\qquad +{\partial ^{2}u \over \partial x_{1}\,\partial x_{2}}\cdot {\partial v \over \partial x_{3}}+{\partial ^{2}u \over \partial x_{1}\,\partial x_{3}}\cdot {\partial v \over \partial x_{2}}+{\partial ^{2}u \over \partial x_{2}\,\partial x_{3}}\cdot {\partial v \over \partial x_{1}}+{\partial ^{3}u \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}\cdot v.\end{aligned}}$

Banach space

Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B : X × Y → Z is a continuous bilinear operator. Then B is differentiable, and its derivative at the point (x,y) in X × Y is the linear map D_(x,y)B : X × Y → Z given by

$(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.$$(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.$

Derivations in abstract algebra

In abstract algebra, the product rule is used to define what is called a derivation, not vice versa.

Vector functions

The product rule extends to scalar multiplication, dot products, and cross products of vector functions.

For scalar multiplication: $(f\cdot {\mathbf {g}})'=f\;'\cdot {\mathbf {g}}+f\cdot {\mathbf {g}}\;'\,$$(f\cdot {\mathbf {g}})'=f\;'\cdot {\mathbf {g}}+f\cdot {\mathbf {g}}\;'\,$

For dot products: $({\mathbf {f}}\cdot {\mathbf {g}})'={\mathbf {f}}\;'\cdot {\mathbf {g}}+{\mathbf {f}}\cdot {\mathbf {g}}\;'\,$$({\mathbf {f}}\cdot {\mathbf {g}})'={\mathbf {f}}\;'\cdot {\mathbf {g}}+{\mathbf {f}}\cdot {\mathbf {g}}\;'\,$

For cross products: $({\mathbf {f}}\times {\mathbf {g}})'={\mathbf {f}}\;'\times {\mathbf {g}}+{\mathbf {f}}\times {\mathbf {g}}\;'\,$$({\mathbf {f}}\times {\mathbf {g}})'={\mathbf {f}}\;'\times {\mathbf {g}}+{\mathbf {f}}\times {\mathbf {g}}\;'\,$

Note: cross products are not commutative, i.e. $(f\times g)'\neq f'\times g+g'\times f$$(f\times g)'\neq f'\times g+g'\times f$, instead products are anticommutative, so it can be written as $(f\times g)'=f'\times g-g'\times f$$(f\times g)'=f'\times g-g'\times f$

Scalar fields

For scalar fields the concept of gradient is the analog of the derivative:

$\nabla (f\cdot g)=\nabla f\cdot g+f\cdot \nabla g\,$$\nabla (f\cdot g)=\nabla f\cdot g+f\cdot \nabla g\,$

Applications

Among the applications of the product rule is a proof that

${d \over dx}x^{n}=nx^{n-1}\,\!$${d \over dx}x^{n}=nx^{n-1}\,\!$

when n is a positive integer (this rule is true even if n is not positive or is not an integer, but the proof of that must rely on other methods). The proof is by mathematical induction on the exponent n. If n = 0 then xⁿ is constant and nx^n − 1 = 0. The rule holds in that case because the derivative of a constant function is 0. If the rule holds for any particular exponent n, then for the next value, n + 1, we have

${\begin{aligned}{d \over dx}x^{n+1}&{}={d \over dx}\left(x^{n}\cdot x\right)\\[12pt]&{}=x{d \over dx}x^{n}+x^{n}{d \over dx}x\qquad {\mbox{(the product rule is used here)}}\\[12pt]&{}=x\left(nx^{n-1}\right)+x^{n}\cdot 1\qquad {\mbox{(the induction hypothesis is used here)}}\\[12pt]&{}=(n+1)x^{n}.\end{aligned}}$${\begin{aligned}{d \over dx}x^{n+1}&{}={d \over dx}\left(x^{n}\cdot x\right)\\[12pt]&{}=x{d \over dx}x^{n}+x^{n}{d \over dx}x\qquad {\mbox{(the product rule is used here)}}\\[12pt]&{}=x\left(nx^{n-1}\right)+x^{n}\cdot 1\qquad {\mbox{(the induction hypothesis is used here)}}\\[12pt]&{}=(n+1)x^{n}.\end{aligned}}$

Therefore if the proposition is true of n, it is true also of n + 1, and therefore all integral n.

Definition of tangent space

Product rule is also used in definition of abstract tangent space of some abstract geometric figure (smooth manifold). This definition we can use if we cannot or wish to not use surrounding ambient space where our chosen geometric figure lives (since there might be no such surrounding space). It uses the fact that it is possible to define derivatives of real-valued functions on that geometric figure at a point p solely with the product rule and that the set of all such derivations in fact forms a vector space that is the desired tangent space.

See also

• Derivation (differential algebra)
• Differential (mathematics)
• General Leibniz rule
• Quotient rule
• Reciprocal rule

References

• Child, J. M. (2008) "The early mathematical manuscripts of Leibniz", Gottfried Wilhelm Leibniz, translated by J. M. Child; page 29, footnote 58.

External links

• Product Rule Practice Problems [Kouba, University of California: Davis]