School:Product rule for differentiation

For the full article, see product rule for differentiation.

Statement for two functions

Statement in multiple versions

The product rule is stated in many versions:

Version type	Statement
specific point, named functions	Suppose $f$ and $g$ are functions of one variable, both of which are differentiable at a real number $x=x_{0}$. Then, the product function $f\cdot g$, defined as $x\mapsto f(x)g(x)$ is also differentiable at $x=x_{0}$, and the derivative at $x_{0}$ is given as follows: $\!{\frac {d}{dx}}[f(x)g(x)]|_{x=x_{0}}=f'(x_{0})g(x_{0})+f(x_{0})g'(x_{0})$ or equivalently: $\!{\frac {d}{dx}}[f(x)g(x)]|_{x=x_{0}}={\frac {d(f(x))}{dx}}|_{x=x_{0}}\cdot g(x_{0})+f(x_{0})\cdot {\frac {d(g(x))}{dx}}|_{x=x_{0}}$
generic point, named functions, point notation	Suppose $f$ and $g$ are functions of one variable. Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side): $\!{\frac {d}{dx}}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$
generic point, named functions, point-free notation	Suppose $f$ and $g$ are functions of one variable. Then, we have the following equality of functions on the domain where the right side expression makes sense (see concept of equality conditional to existence of one side): $\!(f\cdot g)'=(f'\cdot g)+(f\cdot g')$ We could also write this more briefly as: $\!(fg)'=f'g+fg'$ Note that the domain of $(fg)'$ may be strictly larger than the intersection of the domains of $f'$ and $g'$, so the equality need not hold in the sense of equality as functions if we care about the domains of definition.
Pure Leibniz notation using dependent and independent variables	Suppose $u,v$ are variables both of which are functionally dependent on $x$. Then: $\!{\frac {d(uv)}{dx}}=\left({\frac {du}{dx}}\right)v+u{\frac {dv}{dx}}$
In terms of differentials	Suppose $u,v$ are both variables functionally dependent on $x$. Then, $\!d(uv)=v(du)+u(dv)$.

MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a $\{\}_{0}$ subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.

Significance

Computational feasibility significance

Each of the versions has its own computational feasibility significance:

Version type	Significance
specific point, named functions	This tells us that knowledge of the values (in the sense of numerical values) $\!f(x_{0}),g(x_{0}),f'(x_{0}),g'(x_{0})$ at a specific point $x_{0}$ is sufficient to compute the value of $(f\cdot g)'(x_{0})$. For instance, if we are given that $f(1)=5,g(1)=11,f'(1)=4,g'(1)=13$, we obtain that $(f\cdot g)'(1)=4\cdot 11+5\cdot 13=44+65=109$. A note on contrast with the (false) freshman product rule: [SHOW MORE] If the freshman product rule were true, then knowledge of $\!f'(x_{0})$ and $\!g'(x_{0})$ alone (without knowledge of $f(x_{0})$ and $g(x_{0})$) would be sufficient to compute $(f\cdot g)'(x_{0})$. This is not the case.
generic point, named functions	This tells us that knowledge of the general expressions for $f$ and $g$ and the derivatives of $f$ and $g$ is sufficient to compute the general expression for the derivative of $f\cdot g$. See the #Examples section of this page for more examples.

Examples

Nontrivial examples where simple alternate methods exist

Here is a simple trigonometric example:

$\!f(x):=\sin x\cos x$.

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Doing this using the product rule, we get:

$\!f'(x)=\sin 'x\cos x+\sin x\cos 'x=\cos x\cos x+\sin x(-\sin x)=\cos ^{2}x-\sin ^{2}x=\cos(2x)$

where the last step uses a trigonometric identity.

We could also do this by a different method, noting that, by a trigonometric identity:

$\!f(x)={\frac {1}{2}}\sin(2x)$

and now differentiate, to get:

$\!f'(x)={\frac {1}{2}}{\frac {d}{dx}}\sin(2x)={\frac {1}{2}}\cos(2x){\frac {d}{dx}}(2x)={\frac {1}{2}}\cos(2x)(2)=\cos(2x)$

Nontrivial examples where simple alternate methods do not exist

Consider a product of the form:

$\!f(x):=x\sin x$

Using the product rule, we get:

$f'(x)={\frac {dx}{dx}}\sin x+x{\frac {d}{dx}}(\sin x)=1(\sin x)+x\cos x=\sin x+x\cos x$