Product rule
Encyclopedia : P : PR : PRO : Product rule
In mathematics, the product rule of calculus, also called Leibniz's law (see derivation), governs the differentiation of products of differentiable functions.
It may be stated thus:
- [(fg)'=f'g+fg' \,]
- [(uv)=u+v.]
Discovery by Leibniz
Discovery of this rule is credited to Leibniz, 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
[d(uv)\, ] [= (u + du)(v + dv) - uv\, ] [= u(dv) + v(du) + (du)(dv) \,] Since the term (du)(dv) is "negligible" (i.e. at least quadratic in du and dv), Leibniz concluded that
- [d(uv) = (du)v + u(dv) \,]
- [\frac (uv) = \left( \frac \right) v + u \left( \frac \right)]
- [(uv)' = u' v + u v' \,]
Examples
- Suppose one wants to differentiate f(x) = x2 sin(x). By using the product rule, one gets the derivative f
' (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 f(x) is a differentiable function, then cf(x) is also differentiable, and its derivative is (c × f)
' (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 product rule can be used to derive the rule for integration by parts and the quotient rule.
A common error
It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u′)(v′) (Leibniz himself made this error initially); however, it is quite easy to find counterexamples to this. Most simply, take a function f, whose derivative is f '(x). Now that function can also be written as f(x) · 1, since 1 is the identity element for multiplication. Suppose the above-mentioned misconception were true; if so, (u′)(v′) would equal zero. This is true because the derivative of a constant (such as 1) is zero and the product of f '(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:
Suppose
- [f(x) = g(x)h(x) \,]
[f'(x)\!] [= \lim_ \frac] [= \lim_ \frac] [= \lim_ \frac] [= \lim_ \frac] [= \lim_ \left(g(x)\frac + h(x + \Delta x) \frac\right)] Since h is continuous at x, we have
- [\lim_ h(x + \Delta x) = h(x)]
- [g'(x)=\lim_ \frac]
- [h'(x)=\lim_ \frac]
- [f'(x) = \lim_ \left[g(x)left(fracright) + h(x + Delta x)left(fracright)right]]
- [= \left[lim_ g(x)right]\left[lim_ fracright] + \left[lim_ h(x + Delta x)right]\left[lim_fracright]]
- [= g(x)h'(x) + h(x)g'(x) \,]
Generalizations
The product rule can be generalised to products of more than two factors. For example, for three factors we have
- [\frac = \fracvw + u\fracw + uv\frac]
- [\frac \prod_^k f_i(x) = \left(\sum_^k \frac f_i(x)}\right) \prod_^k f_i(x)]
- [y^(x) = \sum_^n u^(x)\; v^(x)].
In multivariable calculus, the product rule is also valid for different notions of "product": scalar product and cross product of vectors, matrix product, inner products etc. All of these are summarized by the following general statement: let X, Y, Z be Banach spaces (which includes Euclidean space) and let B : X × Y → Z be 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( x,y \right) + B_\left( u,v \right)\qquad\forall (u,v)\in X \times Y ].
Derivation in abstract algebra
In abstract algebra, the product rule is used to define what is called a derivation, not vice versa.
See also
From Wikipedia, the Free Encyclopedia. Original article here. Support Wikipedia by contributing or donating.
All text is available under the terms of the GNU Free Documentation License See Wikipedia Copyrights for details.
