Capítulo II Estado de la Cuestión
2.1.2 Encuadres noticiosos hacia el papel de una primera dama
One quickly learns in Calculus that although the derivative is defined as a limit of a difference quotient, there is a small collection of algorithms that reduce the finding of the derivative of any combination of elementary functions to a fairly mechanical exercise. The algorithms show you how to take the derivatives of the sum, difference, product, and quotient of two differentiable functions as well as a constant multiple of a differentiable function, the inverse of a differentiable function, and the composition of two differentiable functions. Those rules along with the knowledge of how to differentiate the elementary functions, xn, ax, logax, sin x, and cos x give you all the tools necessary to differentiate virtually any function you are likely to see in a lifetime of applications. This and the next sections discusses the proofs of the theorems that provide these needed algorithms.
The simplest of these results is the theorem that states that if f is a function differentiable at a and c is any constant, then the function cf is also differentiable at a with.cf /0.a/ D cf0.a/. In the proof of this theorem, you would assume that f0.a/
exists. That provides for you the limit lim
x!a f.x/f .a/
xa D f0.a/. Since the limit needed to show that.cf /0.a/ D cf0.a/ is just a multiple of a known limit, the needed result follows immediately from the fact that the limit of a constant times a function is the constant times the limit of the function.
PROOF: If f0.a/ exists, and c is a constant, then .cf/0.a/ D cf0.a/.
• Suppose that f has a derivative at the point a.
• From the definition of derivative f0.a/ D lim
x!a
To show that the derivative of the sum or difference of two differentiable functions is the sum or difference of their derivatives, one is faced with finding the limit of a difference quotient which can easily be written as the sum or difference of two difference quotients whose limits are already known. Thus, if f and g are two functions defined on the same domain and both differentiable at a, calculating the derivative of f C g at a requires the limit
xlim!a
PROOF: Let f and g be functions defined on a common domain, and let f and g both be differentiable at a. Then.f C g/0.a/ D f0.a/ C g0.a/ and .f g/0.a/ D f0.a/ g0.a/.
• Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a.
• From the definition of derivative f0.a/ D lim
x!a
Why does the first step in this proof make the assumption that f and g are defined on the same domain? This is to avoid the embarrassing situation that the intersection of the domains of f and g isolates the point a. For example, if f is defined for all x 1 and g is defined for all x 1, it could be that both f0.1/ and g0.1/ are defined, but the function f Cg is defined only at 1, so its derivative cannot be defined. Another
138 5 Derivatives example would be for f to be defined at all rational numbers, and g to be defined at all rational multiples ofp
2. Each function could be differentiable at each point of its domain, but f C g is only defined at 0, so its derivative cannot be defined.
It is certainly worth noting here that the theorems discussed so far show that for functions f and g and constants a and b, the derivative of the linear combination of functions af.x/Cbg.x/ is the linear combination of the derivatives af0.x/Cbg0.x/.
In the words of Linear Algebra, this says that the derivative is a linear operator. This fact alone has a long list of ramifications in Differential Equations and other fields.
It is important for the beginning Calculus student to learn that even though the derivative behaves in an intuitive way with respect to addition and subtraction, that this intuition ceases when discussing the derivative of a product or quotient. The proof that.fg/0D fg0C f0g involves one trick reminiscent of the proof that the limit of a product is the product of the limits. That is, one adds and subtracts the same quantity so that rather than making a change in two different factors at the same time, one makes a change in one factor at a time. Indeed, the difference quotient you obtain for the function fg is
f.x/g.x/ f .a/g.a/
Taking the limits at each step produces the following proof.
PROOF (Product Rule): Let f and g be functions defined on a common domain, and let f and g both be differentiable at a. Then .fg/0.a/ D f.a/g0.a/ C f0.a/g.a/.
• Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a.
• From the definition of derivative f0.a/ D lim
x!a
• Because f is differentiable at a, it is continuous at a. This implies that
xlim!af.x/ D f .a/.
D gf0gfg2 0 involves the same strategy along with the extra assumption that g.a/ ¤ 0.
PROOF (Quotient Rule): Let f and g be functions defined on a common
• Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a with g.x/ ¤ 0.
• From the definition of derivative f0.a/ D lim
x!a
• Because g is differentiable at a, it is continuous at a. This implies that
xlim!ag.x/ D g.a/.
Write proofs for each of the following statements.
1. For any constant c, the function f.x/ D c has derivative f0.x/ D 0.
140 5 Derivatives domain and each differentiable at a, the derivative
f1f2f3 fn
The Chain Rule shows how to differentiate a function that is a composition of other functions. Since composition is an invaluable tool for constructing functions, the Chain Rule deserves its place among the important algorithms for calculating derivatives. It states that if g is a function differentiable at a, and if f is a function defined on the range of g and differentiable at g.a/, then the function .f ı g/.x/ is differentiable at a, and .f ı g/0.a/ D f0
g.a/
g0.a/. To prove this, you would need to find the limit of the difference quotient D D f
g.x/
f
g.a/
xa .
Expecting to see the expression g.x/g.a/xa which has limit g0.a/, you might try both multiplying and dividing the difference quotient D by the factor g.x/ g.a/ to get
f
g.x/
f
g.a/
g.x/g.a/ g.x/g.a/xa . This idea leads to the following almost correct proof.
PROOF ATTEMPT: Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/. Then .f ı g/0.a/ D f0
g.a/ g0.a/.
• Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/.
• From the definition of derivative, g0.a/ D lim
x!a
• Because g is differentiable at a, it is continuous at a. This implies that
xlim!ag.x/ D g.a/.