In calculus, the Product Rule (or Leibniz rule[1] 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

{\displaystyle (u\cdot v)'=u'\cdot v+u\cdot v'}

or in Leibniz’s notation as

{\displaystyle {\dfrac {d}{dx}}(u\cdot v)={\dfrac {du}{dx}}\cdot v+u\cdot {\dfrac {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.[2] (However, J. M. Child, a translator of Leibniz’s papers,[3] 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

{\displaystyle 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

{\displaystyle {\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

{\displaystyle (u\cdot v)'=v\cdot u'+u\cdot v'.}

**Examples**

- Suppose we want to differentiate
*f*(*x*) =*x*^{2}sin(*x*). By using the product rule, one gets the derivative*f′*(*x*) = 2*x*sin(*x*) +*x*^{2}cos(*x*) (since the derivative of*x*^{2}is 2*x*and the derivative of the sine function is the cosine function). - One special case of the product rule is 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*) =*c f′*(*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,

{\displaystyle 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.

{\displaystyle {\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

{\displaystyle \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 : R –> R** are differentiable at

**then we can write**

*x*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~~ {\displaystyle \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). Then:

{\displaystyle {\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)+f'(x)g(x)h+f(x)g'(x)h-fg(x)+{\text{other terms}}\\&=f'(x)g(x)h+f(x)g'(x)h+o(h)\\[12pt]\end{aligned}}}

The “other terms” consist of items such as

{\displaystyle f(x)\psi _{2}(h),f'(x)g'(x)h^{2}}~~ and ~~{\displaystyle hf'(x)\psi _{1}(h).}

It is not difficult to show that they are all ** o (h)** Dividing by

**and taking the limit for small**

*h***gives the result.**

*h***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

{\displaystyle q(x)={\tfrac {x^{2}}{4}}}):

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

Differentiating both sides:

{\displaystyle {\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.

{\displaystyle {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

{\displaystyle {\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

{\displaystyle {\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

{\displaystyle du\,dv=u'v'(dx)^{2}=0.}

**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

{\displaystyle {\frac {d(uvw)}{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}.}

For a collection of functions ** f_{1} ,…. , f_{k}** , we have

{\displaystyle {\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

{\displaystyle \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

{\displaystyle \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**

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:

{\displaystyle 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 *n*th derivative of an arbitrary number of factors:

{\displaystyle \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 2^{n} subsets of {1, …, n}, and |S| is the cardinality of S. For example, when n = 3,

{\displaystyle {\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 × 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.

**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.[5]

Scalar multiplication:

{\displaystyle (f\cdot \mathbf {g} )'=f'\cdot \mathbf {g} +f\cdot \mathbf {g} '}

For dot products:

{\displaystyle (\mathbf {f} \cdot \mathbf {g} )'=\mathbf {f} '\cdot \mathbf {g} +\mathbf {f} \cdot \mathbf {g} '}

For cross products:

{\displaystyle (\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:

{\displaystyle \nabla (f\cdot g)=\nabla f\cdot g+f\cdot \nabla g}

**Applications**

Among the applications of the product rule is a proof that

{\displaystyle {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 {{(the~~ product~~ rule~~ is~~ used ~~here)}}\\[12pt]&{}=x\left(nx^{n-1}\right)+x^{n}\cdot 1\qquad {{(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*.