User:Support.and.Defend/product rule

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


 * $$(f\cdot g)'=f'\cdot g+f\cdot g' \,\! $$

or in the Leibniz notation thus:


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

The derivative of the product of three functions is:


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

Since the product of two or more functions occurs in many mathematical models of physical phenomena, the product rule has broad application in Physics, Chemistry, and Engineering.

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



\begin{align} d(u\cdot v) & {} = (u + du)\cdot (v + dv) - u\cdot v \\ & {} = u\cdot dv + v\cdot du + du\cdot dv. \end{align} $$

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 "prime notation" as


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

Examples

 * Suppose one wants to differentiate &fnof;(x) = x2 sin(x). By using the product rule, one gets the derivative &fnof; ' (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 &fnof;(x) is a differentiable function, then c&fnof;(x) is also differentiable, and its derivative is (c &times; &fnof;) ' (x) = c &times; &fnof; ' (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.)

Physics Example I: Rocket Acceleration
Consider the vertical acceleration of a model rocket relative to its initial position at a fixed point on the ground. Newton's second law says that the force is equal to the time rate change of momentum. If F is the net force (sum of forces), p is the momentum, and t is the time,



\begin{align} F = \frac{dp}{dt}. \end{align} $$

Since the momentum is equal to the product of mass and velocity, this yields



\begin{align} F = \frac{d}{dt}\left( mv \right), \end{align} $$ where m is the mass and v is the velocity. Application of the product rule gives



\begin{align} F = v\frac{dm}{dt} + m\frac{dv}{dt}. \end{align} $$

Since the acceleration a, is defined as the time rate change of velocity, a = dv/dt,

\begin{align} F = v\frac{dm}{dt} + ma. \end{align} $$

Solving for the acceleration,

\begin{align} a= \frac{F - v\frac{dm}{dt}}{m}. \end{align} $$

Since the rocket is losing mass, dm/dt is negative, and the changing mass term results in increased acceleration.

Physics Example II: Electromagnetic induction
Faraday's law of electromagnetic induction states that the induced electromotive force is the negative time rate of change of magnetic flux through a conducting loop.

$$ \mathcal{E} = -{{d\Phi_B} \over dt},$$

where $$\mathcal{E}$$ is the electromotive force (emf) in volts and &Phi;B is the magnetic flux in webers. For a loop of area, A, in a magnetic field, B, the magnetic flux is given by

$$ \Phi_B = B\cdot A \cdot \cos(\theta),$$

where θ is the angle between the normal to the current loop and the magnetic field direction.

Taking the negative derivative of the flux with respect to time yields the electromotive force gives

$$ \begin{align} \mathcal{E} &= -\frac{d}{dt} \left( B\cdot A \cdot \cos(\theta) \right) \\ &= -\frac{dB}{dt} \cdot A \cos(\theta) -B \cdot \frac{dA}{dt} \cos(\theta)- B \cdot A \frac{d}{dt}\cos(\theta)\\ \end{align} $$

In many cases of practical interest, only one variable (A, B, or θ) is changing so two of the three above terms are often zero.

A common error
It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u &prime;)(v &prime;) (there is an exaggerated story that Leibniz himself made this error initially); however, there are clear counterexamples to this. For a &fnof;(x) whose derivative is &fnof; '(x), the function can also be written as &fnof;(x) · 1, since 1 is the identity element for multiplication. If the above-mentioned misconception were true, (u&prime;)(v&prime;) would equal zero. This is true because the derivative of a constant (such as 1) is zero and the product of &fnof; ' (x) · 0 is also zero.

Proof of the product rule
A rigorous proof of the product rule can be given using the properties of limits and the definition of the derivative as a limit of Newton's difference quotient.

If


 * $$ h(x) = f(x)g(x),\,$$

and &fnof; and g are each differentiable at the fixed number x, then


 * $$h'(x) = \lim_{w\to x}{ h(w) - h(x) \over w - x} = \lim_{w\to x}{f(w)g(w) - f(x)g(x) \over w - x}. \qquad\qquad(1)$$

Now the difference


 * $$ f(w)g(w) - f(x)g(x)\qquad\qquad(2) $$

is the area of the big rectangle minus the area of the small rectangle in the illustration. The region between the smaller and larger rectangle can be split into two rectangles, the sum of whose areas is


 * $$ f(x) \Bigg( g(w) - g(x) \Bigg) + g(w)\Bigg( f(w) - f(x) \Bigg).\qquad\qquad(3) $$

Therefore the expression in (1) is equal to


 * $$\lim_{w\to x}\left( f(x) \left( {g(w) - g(x) \over w - x} \right) + g(w)\left( {f(w) - f(x) \over w - x} \right) \right).\qquad\qquad(4)$$

Assuming that all limits used exist, (4) is equal to


 * $$ \left(\lim_{w\to x}f(x)\right) \left(\lim_{w\to x} {g(w) - g(x) \over w - x}\right)

+ \left(\lim_{w\to x} g(w)\right) \left(\lim_{w\to x} {f(w) - f(x) \over w - x} \right). \qquad\qquad(5) $$

Now


 * $$\lim_{w\to x}f(x) = f(x)\,$$

because &fnof;(x) remains constant as w → x;


 * $$ \lim_{w\to x} {g(w) - g(x) \over w - x} = g'(x) $$

because g is differentiable at x;


 * $$ \lim_{w\to x} {f(w) - f(x) \over w - x} = f'(x) $$

because &fnof; is differentiable at x;

and now the "hard" one:


 * $$ \lim_{w\to x} g(w) = g(x)\, $$

because g, being differentiable, is continuous at x.

We conclude that the expression in (5) is equal to


 * $$ f(x)g'(x) + g(x)f'(x). \,$$

Alternative proof
Suppose :$$y = f(x)g(x)\,\!$$

By applying Newton's difference quotient and the limit as h approaches 0, we are able to represent the derivative in the form


 * $$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h)g(x + h) - f(x)g(x)}{h}\,\!$$

In order to simplify this limit we add and subtract the term $$f(x)g(x + h)$$ to the numerator, keeping the fraction's value unchanged


 * $$= \lim_{h \to 0} \frac{f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h)}{h}\,\!$$

This allows us to factorise the numerator like so


 * $$= \lim_{h \to 0} \frac{f(x)(g(x + h) - g(x)) + g(x + h)(f(x + h) - f(x))}{h}\,\!$$

The fraction is split into two


 * $$= \lim_{h \to 0} \left ( f(x)\frac{g(x + h) - g(x)}{h} + g(x + h)\frac{f(x + h) - f(x)}{h} \right )\,\!$$

The limit is applied to each term and factor of the limit expression


 * $$= \lim_{h \to 0} f(x) \times \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} + \lim_{h \to 0} g(x + h) \times \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\,\!$$

Each limit is evaluated. Taking into consideration the definition of the derivative, the result is


 * $$= f(x)g'(x) + g(x)f'(x)\,\!$$

Using logarithms
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,


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

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. 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.

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


 * $$ {d (ab) \over dx} = \frac{\partial(ab)}{\partial a}\frac{da}{dx}+\frac{\partial (ab)}{\partial b}\frac{db}{dx} = b \frac{da}{dx} + a \frac{db}{dx}. \, $$

Using non-standard analysis
Let u and v be continuous functions in x, and let dx, du and dv be infinitesimals. This gives,




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

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


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

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(\frac{d}{dx} f_i(x) \prod_{j\ne 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:
 * $$(uv)^{(n)}(x) = \sum_{k=0}^n {n \choose k} \cdot u^{(n-k)}(x)\cdot v^{(k)}(x).$$

See also binomial coefficient and the formally quite similar binomial theorem. See also Leibniz rule (generalized product rule).

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}. If this seems hard to understand, consider the case in which n = 3:


 * $$\begin{align} &{}\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{align}$$

A product rule in Banach spaces
Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B : X &times; Y → Z is a continuous bilinear operator. Then B is differentiable, and its derivative at the point (x,y) in X &times; Y is the linear map D(x,y)B : X &times; 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.

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

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

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

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

(Beware: since cross products are not commutative, it is not correct to write $$(f\times g)'=f'\times g+g'\times f. \,$$ But cross products are anticommutative, so it can be written as $$(f\times g)'=f'\times g-g'\times f. \,$$)

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

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

An application
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 &minus; 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{align}

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

Therefore if the proposition is true of n, it is true also of n + 1.