`http://www.mathsisfun.com/algebra/matrix-multiplying.html`

first.
(This is very elementary.)
Now, if you have *m* functions of *n* variables each,
then you can put their partial derivatives into an *m*-by-*n* matrix;
for example, if you have 2 functions of 3 variables each,
say *u* = *f*(*x*, *y*, *z*)
and *v* = *g*(*x*, *y*, *z*),
then the partial derivatives fit into a 2-by-3 matrix

⎡ | ∂u/∂x |
∂u/∂y |
∂u/∂z | ⎤ |

⎣ | ∂v/∂x |
∂v/∂y |
∂v/∂z | ⎦; |

⎡ | D_{1}f |
D_{2}f |
D_{3}f | ⎤ |

⎣ | D_{1}g |
D_{2}g |
D_{3}g | ⎦, |

If you have an ordinary function *y* = *f*(*x*),
you can think of this as a group of only 1 function of only 1 variable each,
so that d(*y*)/d(*x*) = D(*f*)(*x*)
is a 1-by-1 matrix,
consisting of a single entry,
which is the usual derivative
d*y*/d*x* = *f*′(*x*).
That is, d(*y*)/d(*x*) = [d*y*/d*x*],
and D(*f*) = [*f*′].

If you have a parametrized curve in 3 dimensions,
say *P* = (*x*, *y*, *z*) =
(*f*(*t*), *g*(*t*), *h*(*t*)),
then this is a group of 3 functions of 1 variable each,
so that d(*x*, *y*, *z*)/d(*t*) =
D(*f*, *g*, *h*)(*t*)
is a 3-by-1 matrix, consisting of a single column with 3 entries,
which are the components of the velocity vector
d*P*/d*t* =
⟨*f*′(*t*), *g*′(*t*), *h*′(*t*)⟩.
(For this reason, ordinary vectors that represent change *of* a point
are sometimes called *column vectors*.)

If you have a function of 3 variables,
say *u* = *F*(*x*, *y*, *z*),
you can think of this as a group of 1 function of 3 variables each,
so that d(*u*)/d(*x*, *y*, *z*) =
D(*F*)(*x*, *y*, *z*)
is a 1-by-3 matrix, consisting of a single row with 3 entries,
which are the components of the gradient vector
⟨∂*u*/∂*x*, ∂*u*/∂*y*, ∂*u*/∂*z*⟩ =
∇*F*(*x*, *y*, *z*).
(For this reason,
vectors such as gradients
that represent change *with respect to* a point
are sometimes called *row vectors*.)

If you have both
(*x*, *y*, *z*) =
(*f*(*t*), *g*(*t*), *h*(*t*))
and *u* = *F*(*x*, *y*, *z*),
then composition makes *u* an ordinary function of *t*;
specifically,
*u* =
*F* ∘ (*f*, *g*, *h*).
Recall the defining property of the gradient
from page 26 of my notes:
(*F* ∘ (*f*, *g*, *h*))′(*t*) =
∇*F*(*f*(*t*), *g*(*t*), *h*(*t*)) ·
⟨*f*′(*t*), *g*′(*t*), *h*′(*t*)⟩;
or d*u*/d*t* =
⟨∂*u*/∂*x*, ∂*u*/∂*y*, ∂*u*/∂*z*⟩ ·
⟨d*x*/d*t*, d*y*/d*t*, d*z*/d*t*⟩.
The same thing can be expressed using matrix multiplication as
d(*u*)/d(*t*) =
d(*u*)/d(*x*, *y*, *z*)
d(*x*, *y*, *z*)/d(*t*),
because a matrix row by a matrix column are multiplied
by the same method as the dot product.

More generally, all of the forms of the Chain Rule
in Section 13.4 of the textbook
can be expressed using matrix multiplication.
If a point *R* (in any number of dimensions)
is a function of a point *Q* (in any number of dimensions),
which is a function of a point *P* (in any number of dimensions),
then d*R*/d*P* = d*R*/d*Q* d*Q*/d*P*.

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This web page was written between 2003 and 2018 by Toby Bartels, last edited on 2018 April 17. Toby reserves no legal rights to it.

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