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Difference between revisions of "Product rule of differentiation"
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Latest revision as of 21:27, 28 August 2011
For vector functions:
Theorem (Product Rule). Suppose
Then
Note that since $f(a)$ is of dimension $m \times 1$, $k'(a)$ is of dimension $1 \times n$ (and hence $f(a) k'(a)$ is of dimension $m \times n$),
$k(a)$ is a scalar, $f'(a)$ is of dimension $m \times n$ (and hence $k(a) f'(a)$ is of dimension $m \times n$), the derivative $( kf )'(a)$ is a matrix of dimension $m \times n$.
Further keep in mind that here
- derivatives are matrices,
- functions are vectors.