 Geometric illustration of a proof of the product rule

In calculus, the product rule (or Leibniz rule or Leibniz product rule) is a formula used to find the derivatives of products of two or more functions. For two functions, it may be stated in Lagrange's notation as

$(u\cdot v)'=u'\cdot v+u\cdot v'$ or in Leibniz's notation as
${\frac {d}{dx))(u\cdot v)={\frac {du}{dx))\cdot v+u\cdot {\frac {dv}{dx)).$ The rule may be extended or generalized to products of three or more functions, to a rule for higher-order derivatives of a product, and to other contexts.

## Discovery

Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials. (However, J. M. Child, a translator of Leibniz's papers, 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 f(x) = x2 sin(x). By using the product rule, one gets the derivative f(x) = 2x sin(x) + x2 cos(x) (since the derivative of x2 is 2x and the derivative of the sine function is the cosine function).
• One special case of the product rule is the constant multiple rule, which states: if c is a number and f(x) is a differentiable function, then cf(x) is also differentiable, and its derivative is (cf)(x) = cf(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 (from first principles)

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, $f(x)g(x+\Delta x)-f(x)g(x+\Delta x)$ (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.

{\begin{aligned}h'(x)&=\lim _{\Delta x\to 0}{\frac {h(x+\Delta x)-h(x)}{\Delta x))\\[5pt]&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x))\\[5pt]&=\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))\\[5pt]&=\lim _{\Delta x\to 0}{\frac ((\big [}f(x+\Delta x)-f(x){\big ]}\cdot g(x+\Delta x)+f(x)\cdot {\big [}g(x+\Delta x)-g(x){\big ])){\Delta x))\\[5pt]&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x))\cdot \underbrace {\lim _{\Delta x\to 0}g(x+\Delta x)} _{\text{See the note below.))+\lim _{\Delta x\to 0}f(x)\cdot \lim _{\Delta x\to 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x))\\[5pt]&=f'(x)g(x)+f(x)g'(x).\end{aligned)) The fact that

$\lim _{\Delta x\to 0}g(x+\Delta x)=g(x)$ is deduced from a theorem that states that differentiable functions are continuous.

### Brief proof

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

$f(x+h)=f(x)+f'(x)h+\psi _{1}(h)$ and
$g(x+h)=g(x)+g'(x)h+\psi _{2}(h)$ such that ${\textstyle \lim _{h\to 0}{\frac {\psi _{1}(h)}{h))=\lim _{h\to 0}{\frac {\psi _{2}(h)}{h))=0,}$ also written $\psi _{1},\psi _{2}\sim o(h^{3})$ . Then:
{\begin{aligned}f(x+h)g(x+h)-f(x)g(x)&=(f(x)+f'(x)h+\psi _{1}(h))(g(x)+g'(x)h+\psi _{2}(h))-f(x)g(x)\\&=f(x)g(x)+f'(x)g(x)h+f(x)g'(x)h-f(x)g(x)+{\text{other terms))\\&=f'(x)g(x)h+f(x)g'(x)h+o(h^{3})\\[12pt]\end{aligned)) The "other terms" consist of items such as $f(x)\psi _{2}(h),f'(x)g'(x)h^{2)$ and $hf'(x)\psi _{1}(h).$ It is not difficult to show that they are all $o(h^{3}).$ Dividing by $h$ and taking the limit for small $h$ gives the result.

### Quarter squares

There is a proof using quarter square multiplication which relies on the chain rule and on the properties of the quarter square function (shown here as q, i.e., with $q(x)={\tfrac {x^{2)){4))$ ).

Let

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

{\begin{aligned}f'&=q'(u+v)(u'+v')-q'(u-v)(u'-v')\\[4pt]&=\left({1 \over 2}(u+v)(u'+v')\right)-\left({1 \over 2}(u-v)(u'-v')\right)\\[4pt]&={1 \over 2}(uu'+vu'+uv'+vv')-{1 \over 2}(uu'-vu'-uv'+vv')\\[4pt]&=vu'+uv'\\[4pt]&=uv'+u'v\end{aligned)) ### 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

{\begin{aligned}{\frac {d(uv)}{dx))&=\operatorname {st} \left({\frac {(u+du)(v+dv)-uv}{dx))\right)\\[4pt]&=\operatorname {st} \left({\frac {uv+u\cdot dv+v\cdot du+dv\cdot du-uv}{dx))\right)\\[4pt]&=\operatorname {st} \left({\frac {u\cdot dv+(v+dv)\cdot du}{dx))\right)\\[4pt]&=u{\frac {dv}{dx))+v{\frac {du}{dx)).\end{aligned)) 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

$du\,dv=u'v'(dx)^{2}=0.$ ### Using log

Let $f(x)=u(x)v(x)$ Taking natural log on both sides,

$\ln f=\ln u+\ln v$ Differentiating wrt x,

{\begin{aligned}{\dfrac {f'}{f))&={\dfrac {u'}{u))+{\dfrac {v'}{v))\\f'&=uv({\dfrac {u'}{u))+{\dfrac {v'}{v)))\\f'&=u'v+uv'\end{aligned)) ## Generalizations

### 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

${\frac {d(uvw)}{dx))={\frac {du}{dx))vw+u{\frac {dv}{dx))w+uv{\frac {dw}{dx)).$ For a collection of functions $f_{1},\dots ,f_{k)$ , we have

${\frac {d}{dx))\left[\prod _{i=1}^{k}f_{i}(x)\right]=\sum _{i=1}^{k}\left(\left({\frac {d}{dx))f_{i}(x)\right)\prod _{j=1,j\neq i}^{k}f_{j}(x)\right)=\left(\prod _{i=1}^{k}f_{i}(x)\right)\left(\sum _{i=1}^{k}{\frac {f'_{i}(x)}{f_{i}(x)))\right).$ The logarithmic derivative provides a simpler expression of the last form, as well as a direct proof that does not involve any recursion. The logarithmic derivative of a function f, denoted here Logder(f), is the derivative of the logarithm of the function. It follows that

$\operatorname {Logder} (f)={\frac {f'}{f)).$ Using that the logarithm of a product is the sum of the logarithms of the factors, the sum rule for derivatives gives immediately

$\operatorname {Logder} (f_{1}\cdots f_{k})=\sum _{i=1}^{k}\operatorname {Logder} (f_{i}).$ The last above expression of the derivative of a product is obtained by multiplying both members of this equation by the product of the $f_{i}.$ ### Higher derivatives

 Main article: General Leibniz rule

It can also be generalized to the general Leibniz rule for the nth derivative of a product of two factors, by symbolically expanding according to the binomial theorem:

$d^{n}(uv)=\sum _{k=0}^{n}{n \choose k}\cdot d^{(n-k)}(u)\cdot d^{(k)}(v).$ Applied at a specific point x, the above formula gives:

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

$\left(\prod _{i=1}^{k}f_{i}\right)^{(n)}=\sum _{j_{1}+j_{2}+\cdots +j_{k}=n}{n \choose j_{1},j_{2},\ldots ,j_{k))\prod _{i=1}^{k}f_{i}^{(j_{i})}.$ ### 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 all 2n subsets of {1, ..., n}, and |S| is the cardinality of S. For example, when n = 3,

{\begin{aligned}&{\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3))(uv)\\[6pt]={}&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))\\[6pt]&+{\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 × YZ 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 × YZ given by

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

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

### In vector calculus

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

• For scalar multiplication:
$(f\cdot \mathbf {g} )'=f'\cdot \mathbf {g} +f\cdot \mathbf {g} '$ • For dot products:
$(\mathbf {f} \cdot \mathbf {g} )'=\mathbf {f} '\cdot \mathbf {g} +\mathbf {f} \cdot \mathbf {g} '$ • For cross products:
$(\mathbf {f} \times \mathbf {g} )'=\mathbf {f} '\times \mathbf {g} +\mathbf {f} \times \mathbf {g} '$ There are also analogues for other analogs of the derivative: if f and g are scalar fields then there is a product rule with the gradient:

$\nabla (f\cdot g)=\nabla f\cdot g+f\cdot \nabla g$ ## Applications

Among the applications of the product rule is a proof that

${d \over dx}x^{n}=nx^{n-1)$ 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 nxn − 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)) Therefore, if the proposition is true for n, it is true also for n + 1, and therefore for all natural n.