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Redefining the derivative
Matrices appear in many different contexts in mathematics, not just when we need to solve a system of linear equations. An important instance is linear approximation. Recall from your calculus course that a differentiable function f can be expanded about any point a in its domain using Taylor's theorem. We can write
where c is some point between x and a. The remainder term is the "error"made by using the linear approximation to f at x = a,
That is, f (x) minus the approximation is exactly equal to the error (remainder) term. In fact, we can write Taylor's theorem in the more suggestive form
f (x) = f (a) + f '(a)(x a) + ∈(x, a),
where the remainder term has now been renamed the error term (x; a) and has the important property
(The existence of this limit is another way of saying that the error term \looks like" (x  a)^{2}.) This observation gives us an alternative (and in fact, much better) de nition of the derivative:
De nition: The realvalued function f is said to be differentiable at x = a if there exists a number A and a function ∈(x, a) such that
f (x) = f (a) + A(x  a) + ∈(x, a),
where
Remark: the error term clearly depends on a, and it depends on x as well since the number c varies with x.
Theorem: This is equivalent to the usual calculus definition.
Proof: If the new de nition holds, then if we compute f '(x) by the usual de nition, we find
and A = f '(a) according to the standard de nition. Conversely, if the standard de nition of differentiability holds, then we can de ne ∈(x, a) to be the error made in the linear
approximation:
∈(x, a) = f (x)  f (a)  f '(a)(x  a)
Then
so f can be written in the new form, with A = f '(a).
Example: Let f (x) = 4 + 2x  x^{2}, and let a = 2. So f (a) = f (2) = 4; and f '(a) = f '(2) = 2  2a =  2.. Now subtract f (2) + f '(2)(x  2) from f (x) to get
4 + 2x  x^{2} (4  2(x  2)) = 4 + 4x  x^{2} = (x  2)^{2}.
This is the error term, which is quadratic in x  2, as advertised. So 8  2x(= f (2) + f '(2)(x  2)) is the correct linear approximation to f at x = 2.
Suppose we try some other linear approximation  for example, we could try f (2)  4(x  2) = 12  4x: Subtracting this from f (x) gives 8 + 6x  x^{2} = 2(x  2)  (x  2)^{2}, which is our new error term. But this won't work, since
which is clearly not 0. The only \linear approximation" that leaves a quadratic remainder as the error term is the one formed in the usual way, using the derivative.
Exercise: Interpret this geometrically in terms of the slope of various lines passing through the point (2, f (2)).
Generalization to higher dimensions
Our new de nition of derivative is the one which generalizes to higher dimensions. We start with an
Example: Consider a function from R^{2} to R^{2}, say
By inspection, as it were, we can separate the right hand side into three parts. We have
and the linear part of f is the vector
which can be written in matrix form as
By analogy with the onedimensional case, we might guess that
f (x) = f (0) + Ax + an error term of order 2 in x; y:
where A is the matrix
And this suggests the following
De nition: A function f : R^{n} → R^{m} is said to be differentiable at the point x = a ∈ R^{n} if there exists an m x n matrix A and a function ∈(x, a) such that
f (x) = f (a) + A(x  a) + ∈(x, a),
where
The matrix A is called the derivative of f at x = a, and is denoted by Df (a).
Generalizing the onedimensional case, it can be shown that if
is differentiable at x = a, then the derivative of f is given by the m x n matrix of partial derivatives
Conversely, if all the indicated partial derivatives exist and are continuous at x = a, then the approximation
is accurate to the second order in x a.
Exercise: Find the derivative of the function f : R^{2} → R^{3} at a = (1, 2)^{t}, where
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