Differentiability of Composite Functions
Exploring the Relationship between Composite Functions and Differentiability
by Preety*, Rekha Rani,
- Published in Journal of Advances in Science and Technology, E-ISSN: 2230-9659
Volume 5, Issue No. 9, May 2013, Pages 0 - 0 (0)
Published by: Ignited Minds Journals
ABSTRACT
Compositefunctions are so common that we usually don't think to think to label them ascomposite functions. However, they arise any time a change in one quantityproduces a change in another which, in turn, produces a change in a thirdquantity. Does that sound confusing? Don't worry, an example will make thingsclear. An Example - For this example, we'll assume that the number of humansliving on the coast affects the number of whales in nearby coastal waters.Since whales eat plankton, the number of whales will affect the number ofplankton in the waters. Let'sbe more specific. Since whales don't like all the noise that people make, theymove out of an area when too many people move in. If we denote the number ofthousands of people by x and the number of whales by y, a simple model would bethat. Nowsince the whales are eating the plankton, more whales mean less plankton. If wemeasure the amount of plankton by z, then a simple model is that. Nowthe end result is that the number of people influence the number of whaleswhich influences the number of plankton. To see how the number of planktondepend on the number of people, we can compute that More generally, if we havetwo functions and , we call the newfunction the composite of and and denote it by .
KEYWORD
composite functions, differentiability, change, quantity, humans, whales, plankton, model, influence, compute
INTRODUCTION
Let f,g,h be continuous real functions such that: where h′ denotes the derivative of h. Using the Dx notation: This is often informally referred to as the chain rule (for differentiation). Leibniz's notation for derivatives (dydx) allows for a particularly elegant statement of this rule: dydx=dydu⋅dudx where:
- dydx is the derivative of y with respect to x
- dydu is the derivative of y with respect to u
- dudx is the derivative of u with respect to x
However, this must not be interpreted to mean that derivatives can be treated as fractions. It simply is a convenient notation.
Corollary
dy/dx=(dydu)(dxdu) for dxdu≠0.
Proof
2
Thus: There are two cases to consider:
Case 1
Suppose g′(x)≠0 and that δx is small but non-zero. Then δy≠0 from (1) above, and: hence the result.
Case 2
Now suppose g′(x)=0 and that δx is small but non-zero. Again, there are two possibilities:
Case 2a
If δy=0, then h(x+δx)−h(x)δx=0. Hence the result.
Case 2b
If δy≠0, then h(x+δx)−h(x)δx=f(y+δy)−f(y)δyδyδx. As δy→0: (1):f(y+δy)−f(y)δy→f′(y) (2):δyδx→0 Thus: limδx→0h(x+δx)−h(x)δx→0=f′(y)g′(x) Again, hence the result. or fog(x) is called a composite function or a function of a function. The theorem for finding the derivative of a composite function is known as the CHAIN RULE.
Theorem :
If f and g are differentiable and are defined by y = f (u) and u = g (x), then the composite function y = f [ g (x) ] is differentiable and we have Corollary : If y = f (u), u = g (v) and v = h (x) where f, g and h are differentiable functions of u, v and x respectively, then
IMPORTANCE OF COMPOSITE FUNCTION
Composite functions are much more common than you may realize. For instance, if you want to compute on your hand-held calculator, you will enter 1.1 and then press the button which squares the entry. After that, you will press the button which exponentiates the entry. Each of the buttons you press, to square and to exponentiate, represent a function. By feeding what comes out of the squaring function into the exponentiation function, you are really computing the composite of these two functions. Said in mathematical notation, you are first evaluating at 1.1 and then putting the result into the function . By composing them, you obtain which you evaluate at 1.1. If we have two functions f(x) and g(x) we can define a composite function h(x)f(g(x)) Thus if f(x)=x3 and g(x)=2x−1 we have h(x)=(2x−1)3=8x3−12x2+6x−1 On the other hand if we define the composite function k(x)g(f(x)) we then have k(x)=2(x3)−1 Notice that h(x) and k(x) are different functions: when composing functions, the order matters. (If you think of functions as procedures, this makes intuitive sense: putting on your socks and then your shoes has a different result from doing this the other way round\dots). In compounding functions such as h(x)=f(g(x)) you have to be a bit careful to ensure that the range of g
Preety1 Rekha Rani2
Composite functions are what you get when you take the output of one function and use it for the input of the next one. In this discussion, we will discuss the composition of functions which are R1 R1, i.e. a real number as an input and a real number as an output. The notation for this is (fg)(x)=f(g(x)), where the output of g(x) becomes the input of f(x) and is described as (fg)(x). As a real example, let's use f(x)=x2+2*x-2, and g(x)=3*x+2. By replacing all of the occurrences of x in f(x) by the formula for g(x) we can find the formula in x for the composite function. So: (fg)(x) = f(g(x)) = f(3*x+2) (fg)(x) = (3*x+2)2+2*(3*x+2)-2 (fg)(x) = (9*x2+12*x+4)+(6*x+4)-2 (fg)(x) = 9*x2+18*x+6 We can also find (gf)(x): (gf)(x) = g(f(x)) = g(x2+2*x-2) (gf)(x) = 3*(x2+2*x-2)+2 (gf)(x) = 3*x2+6*x-4 That's basic composite functions. Compositing functions can do weird things to the domain and range, though, because your doing mappings onto mappings (i.e. mapping one domain to another domain to a range). One important point is that the composite of a function with its inverse yields an identity function. That is f(f-1(x))=f-1(f(x))=x. What's happening is that your mapping the original function to its range and then straight back into the domain.
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