Derivative as a Linear Transformation - Differential Calculus, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

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 real-valued 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 - x², 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 - x² (4 - 2(x - 2)) = -4 + 4x - x² = (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 - x² = -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 R² to R², 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 one-dimensional 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 : Rⁿ → R^m is said to be differentiable at the point x = a ∈ Rⁿ 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 one-dimensional 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 : R² → R³ at a = (1, 2)^t, where