Product rule

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 ${\displaystyle \lim _{\Delta x\to 0}g(x+\Delta x)=g(x)}$ follows from the fact that differentiable functions are continuous.

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

$${\displaystyle f(x+h)=f(x)+f'(x)h+\varepsilon _{1}(h)}$$

and

$${\displaystyle g(x+h)=g(x)+g'(x)h+\varepsilon _{2}(h),}$$

where the error terms are small with respect to h: that is, ${\textstyle \lim _{h\to 0}{\frac {\varepsilon _{1}(h)}{h}}=\lim _{h\to 0}{\frac {\varepsilon _{2}(h)}{h}}=0,}$ also written ${\displaystyle \varepsilon _{1},\varepsilon _{2}\sim o(h)}$. Then:

$${\displaystyle {\begin{aligned}f(x+h)g(x+h)-f(x)g(x)&=(f(x)+f'(x)h+\varepsilon _{1}(h))(g(x)+g'(x)h+\varepsilon _{2}(h))-f(x)g(x)\\[.5em]&=f(x)g(x)+f'(x)g(x)h+f(x)g'(x)h-f(x)g(x)+{\text{error terms}}\\[.5em]&=f'(x)g(x)h+f(x)g'(x)h+o(h).\end{aligned}}}$$

The "error terms" consist of items such as ${\displaystyle f(x)\varepsilon _{2}(h),f'(x)g'(x)h^{2}}$ and ${\displaystyle hf'(x)\varepsilon _{1}(h)}$ which are easily seen to have magnitude ${\displaystyle o(h).}$ Dividing by ${\displaystyle h}$ and taking the limit ${\displaystyle h\to 0}$ gives the result.

This proof uses the chain rule and the quarter square function ${\displaystyle q(x)={\tfrac {1}{4}}x^{2}}$ with derivative ${\displaystyle q'(x)={\tfrac {1}{2}}x}$. We have:

${\displaystyle uv=q(u+v)-q(u-v),}$

and differentiating both sides gives:

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

The product rule can be considered a special case of the chain rule for several variables, applied to the multiplication function ${\displaystyle m(u,v)=uv}$:

${\displaystyle {d(uv) \over dx}={\frac {\partial (uv)}{\partial u}}{\frac {du}{dx}}+{\frac {\partial (uv)}{\partial v}}{\frac {dv}{dx}}=v{\frac {du}{dx}}+u{\frac {dv}{dx}}.}$

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

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

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

In the context of Lawvere's approach to infinitesimals, let ${\displaystyle dx}$ be a nilsquare infinitesimal. Then ${\displaystyle du=u'\ dx}$ and ${\displaystyle dv=v'\ dx}$, 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 ${\displaystyle du\,dv=u'v'(dx)^{2}=0.}$ Dividing by ${\displaystyle dx}$ then gives ${\displaystyle {\frac {d(uv)}{dx}}=u{\frac {dv}{dx}}+v{\frac {du}{dx}}}$ or ${\displaystyle (uv)'=u\cdot v'+v\cdot u'}$.

Let ${\displaystyle h(x)=f(x)g(x)}$. Taking the absolute value of each function and the natural log of both sides of the equation,

${\displaystyle \ln |h(x)|=\ln |f(x)g(x)|}$

Applying properties of the absolute value and logarithms,

${\displaystyle \ln |h(x)|=\ln |f(x)|+\ln |g(x)|}$

Taking the logarithmic derivative of both sides and then solving for ${\displaystyle h'(x)}$:

${\displaystyle {\frac {h'(x)}{h(x)}}={\frac {f'(x)}{f(x)}}+{\frac {g'(x)}{g(x)}}}$

Solving for ${\displaystyle h'(x)}$ and substituting back ${\displaystyle f(x)g(x)}$ for ${\displaystyle h(x)}$ gives:

${\displaystyle {\begin{aligned}h'(x)&=h(x)\left({\frac {f'(x)}{f(x)}}+{\frac {g'(x)}{g(x)}}\right)\\&=f(x)g(x)\left({\frac {f'(x)}{f(x)}}+{\frac {g'(x)}{g(x)}}\right)\\&=f'(x)g(x)+f(x)g'(x).\end{aligned}}}$

Note: Taking the absolute value of the functions is necessary to allow logarithmic differentiation of functions that can have negative values, as logarithms are only defined for positive arguments. This works because ${\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}$, which justifies taking the absolute value of the functions for logarithmic differentiation.

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 ${\displaystyle f_{1},\dots ,f_{k}}$, 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 ${\displaystyle f_{i}.}$

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:

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

Furthermore, for the nth derivative of an arbitrary number of factors, one has a similar formula with multinomial coefficients:

${\displaystyle \left(\prod _{i=1}^{k}f_{i}\right)^{\!\!(n)}=\sum _{j_{1}+j_{2}+\cdots +j_{k}=n}{n \choose j_{1},j_{2},\ldots ,j_{k}}\prod _{i=1}^{k}f_{i}^{(j_{i})}.}$

For partial derivatives, we have[4]

${\displaystyle {\partial ^{n} \over \partial x_{1}\,\cdots \,\partial x_{n}}(uv)=\sum _{S}{\partial ^{|S|}u \over \prod _{i\in S}\partial x_{i}}\cdot {\partial ^{n-|S|}v \over \prod _{i\not \in S}\partial x_{i}}}$

where the index S runs through all 2ⁿ subsets of {1, ..., n}, and |S| is the cardinality of S. For example, when n = 3,

${\displaystyle {\begin{aligned}&{\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}(uv)\\[6pt]={}&u\cdot {\partial ^{3}v \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{1}}\cdot {\partial ^{2}v \over \partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{2}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{3}}+{\partial u \over \partial x_{3}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{2}}\\[6pt]&+{\partial ^{2}u \over \partial x_{1}\,\partial x_{2}}\cdot {\partial v \over \partial x_{3}}+{\partial ^{2}u \over \partial x_{1}\,\partial x_{3}}\cdot {\partial v \over \partial x_{2}}+{\partial ^{2}u \over \partial x_{2}\,\partial x_{3}}\cdot {\partial v \over \partial x_{1}}+{\partial ^{3}u \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}\cdot v.\end{aligned}}}$

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

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

This result can be extended[5] to more general topological vector spaces.

The product rule extends to various product operations of vector functions on ${\displaystyle \mathbb {R} ^{n}}$:[6]

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

  
• For dot product:
  $${\displaystyle (\mathbf {f} \cdot \mathbf {g} )'=\mathbf {f} '\cdot \mathbf {g} +\mathbf {f} \cdot \mathbf {g} '}$$

  
• For cross product of vector functions on ${\displaystyle \mathbb {R} ^{3}}$:
  $${\displaystyle (\mathbf {f} \times \mathbf {g} )'=\mathbf {f} '\times \mathbf {g} +\mathbf {f} \times \mathbf {g} '}$$

  

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:

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

Such a rule will hold for any continuous bilinear product operation. Let B : X × Y → Z be a continuous bilinear map between vector spaces, and let f and g be differentiable functions into X and Y, respectively. The only properties of multiplication used in the proof using the limit definition of derivative is that multiplication is continuous and bilinear. So for any continuous bilinear operation,

$${\displaystyle H(f,g)'=H(f',g)+H(f,g').}$$

This is also a special case of the product rule for bilinear maps in Banach space.

In abstract algebra, the product rule is the defining property of a derivation. In this terminology, the product rule states that the derivative operator is a derivation on functions.

In differential geometry, a tangent vector to a manifold M at a point p may be defined abstractly as an operator on real-valued functions which behaves like a directional derivative at p: that is, a linear functional v which is a derivation,

${\displaystyle v(fg)=v(f)\,g(p)+f(p)\,v(g).}$

Generalizing (and dualizing) the formulas of vector calculus to an n-dimensional manifold M, one may take differential forms of degrees k and l, denoted ${\displaystyle \alpha \in \Omega ^{k}(M),\beta \in \Omega ^{\ell }(M)}$, with the wedge or exterior product operation ${\displaystyle \alpha \wedge \beta \in \Omega ^{k+\ell }(M)}$, as well as the exterior derivative ${\displaystyle d:\Omega ^{m}(M)\to \Omega ^{m+1}(M)}$. Then one has the graded Leibniz rule:

${\displaystyle d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1)^{k}\alpha \wedge d\beta .}$