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Unformatted text preview: Growth Authors: Bill Davis, Horacio Porta and Jerry Uhl 1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 1.04 Rules of the Derivative BASICS B.1) Derivatives, instantaneous growth rates, f @ x D and D @ f @ x D , x D B.1.a) What is the derivative of a function f @ x D ? How do you use Mathematica to calculate derivatives? Answer: The derivative f' @ x D of f @ x D is another function that measures the instantaneous growth rate of f @ x D at each point x . Thus "derivative" is just a short-hand way of saying "instantaneous growth rate." Mathematicians have programmed Mathematica to find derivatives of most common functions. Let's see how Mathematica handles some simple functions. To get the derivative of f @ x D = a x 2 + b x + c with respect to x , you have several options: Define: Clear @ f, x, a, b, c D ; f @ x _ D = a x 2 + b x + c c + b x + a x 2 One way to evaluate the derivative with respect to x is to look at: Clear @ h D ; Factor B f @ x + h D- f @ x D h F b + a h + 2 a x As h closes in on , you see that f @ x + h D- f @ x D h closes in on f' @ x D = 2 a x + b . To get the same result from Mathematica , simply type: f @ x D b + 2 a x It works! Or you can type: D @ f @ x D , x D b + 2 a x Chalk up another mark for Mathematica . You can also eliminate the need to define f @ x D : D A a x 2 + b x + c, x E b + 2 a x Another way of doing this is to type: y = a x 2 + b x + c c + b x + a x 2 The command: y I c + b x + a x 2 M is not successful. But you can use: D @ y, x D b + 2 a x This is reminiscent of the notation dy dx that many folks like to use for the derivative of y with respect to x . B.1.b) The D @ , x D notation is particularly useful when the expression you want to differentiate contains parameters or other independent variables. For example, if you are given: Clear @ a, h, f, x, y, z D ; f @ x _ , z _ D = a x 2 + 5 x Sin @ z D- 3- 3 + a x 2 + 5 x Sin @ z D Factor B f @ x + h, z D- f @ x, z D h F a h + 2 a x + 5 Sin @ z D As h closes in on , you see that f @ x + h,z D- f @ x,z D h closes in on the derivative with respect to x a 0 + 2 a x + 5 Sin @ z D = 2 a x + 5 Sin @ z D . You can use Mathematica to get a formula for this derivative with respect to x : D @ f @ x, z D , x D 2 a x + 5 Sin @ z D Here you think of z as a second variable and think of a as a parameter. This distinction is not God-given. There is nothing but custom preventing you from thinking of a as another variable and z as a parameter. In fact, you could think of both as parameters or we could think of both as new independent variables. Here are two calculations for specific values of z : D B f B x, p 8 F , x F 2 a x + 5 Sin B p 8 F D B f B x, p 4 F , x F 5 2 + 2 a x You can calculate the derivative at a particular x : D B f B x, p 4 F , x F . x 8 5 2 + 16 a This is not the same as: D B f B 8, p 4 F , x F Why?...
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- Spring '08