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In Mathematics , the product rule of Calculus , which is also called '''Leibniz's law''' (see Derivation ), governs the Differentiation of products of differentiable Function s.

It may be stated thus:

:(fg)'=f'g+fg' \,

or in the Leibniz notation thus:

:{d\over dx}(uv)=u{dv\over dx}+v{du\over dx}.


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

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) \,

and this is indeed the differential form of the product rule. If we divide through by the differential ''dx'', we obtain

: rac{d}{dx} (uv) = \left( rac{du}{dx} ight) v + u \left( rac{dv}{dx} ight)

which can also be written in "prime notation" as

:(uv)' = u' v + u v' \,


EXAMPLES


  • Suppose you want to differentiate ''f''(''x'') = ''x''2 Sin (''x''). By using the product rule, you get the derivative ''f'''(''x'') = 2''x'' sin(''x'') + ''x''2cos(''x'') (since the derivative of ''x''2 is 2''x'' 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 .



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 Counterexample s 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) \,

and suppose further that ''g'' and ''h'' are each differentiable at the fixed number ''x''. Then
:f'(x) = \lim_{\Delta x o 0} rac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x o 0} rac{g(x + \Delta x)h(x + \Delta x) - g(x)h(x)}{\Delta x}

Since

:g(x + \Delta x)h(x + \Delta x) - g(x)h(x) = g(x)(h(x + \Delta x) - h(x)) + h(x + \Delta x)(g(x + \Delta x) - g(x)),
\,

we have

:f'(x) = \lim_{\Delta x o 0} rac{g(x)(h(x + \Delta x) - h(x)) + h(x + \Delta x)(g(x + \Delta x) - g(x))}{\Delta x}

:= \lim_{\Delta x o 0} \left[g(x)\left( rac{(h(x + \Delta x) - h(x))}{\Delta x} ight) + h(x + \Delta x)\left( rac{(g(x + \Delta x) - g(x))}{\Delta x} ight) ight]

Since ''h'' is continuous at ''x'', we have

:\lim_{\Delta x o 0} h(x + \Delta x) = h(x)

and by the definition of the derivative, and the differentiability of ''h'' and ''g'' at ''x'', we also have

:h'(x)=\lim_{\Delta x o 0} rac{(h(x + \Delta x) - h(x))}{\Delta x} \mbox{ and } g'(x)=\lim_{\Delta x o 0} rac{(g(x + \Delta x) - g(x))}{\Delta x}

Thus, we are justified in splitting each of the products inside the limit, and putting everything together, we have

:f'(x) = \lim_{\Delta x o 0} \left[g(x)\left( rac{(h(x + \Delta x) - h(x))}{\Delta x} ight) + h(x + \Delta x)\left( rac{(g(x + \Delta x) - g(x))}{\Delta x} ight) ight]

:= \left x o 0} g(x) ight \left x o 0} rac{(h(x + \Delta x) - h(x))}{\Delta x} ight + \left x o 0} h(x + \Delta x) ight \left x o 0} rac{(g(x + \Delta x) - g(x))}{\Delta x} ight

:= g(x)h'(x) + h(x)g'(x) \,

and this completes the proof.


GENERALIZATIONS


The product rule can be generalised to products of more than two factors. For example, for three factors we have

: rac{d(uvw)}{dx} = rac{du}{dx}vw + u rac{dv}{dx}w + uv rac{dw}{dx}

For a collection of functions f_1 \dots f_k, we can write this more succinctly as

: rac{d}{dx} \prod_{i=1}^k f_i(x)
= \left(\sum_{i=1}^k rac{ rac{d}{dx} f_i(x)}{f_i(x)} ight)
\prod_{i=1}^k f_i(x)

It can also be generalized to Leibniz Rule for higher derivatives of a product of two factors: if ''y'' = ''uv'' and ''y''(''n'') denotes the ''n''-th derivative of ''y'', then

:y^{(n)}(x) = \sum_{k=0}^n {n \choose k} u^{(n-k)}(x)\; v^{(k)}(x) ,

see also Binomial Coefficient and the formally quite similar Binomial Theorem .

In 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 ight)\,B_\left( u,v ight) = B_\left( x,y ight) + B_\left( u,v ight)\qquad orall (u,v)\in X imes 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