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EXAMPLE 10
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f[x_] = x5 + x4 + x3 + x2 + x + 1; Derivative[1][f] 1 + 2#1 + 3#1 2 + 4#1 3 + 5#1 4 & Derivative[1][f][x] 1 + 2x + 3x 2 + 4x 3 + 5x 4
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Mathematica returns a pure function representing the derivative of f. Pure functions are discussed in the appendix. #1 is replaced by x.
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The numerical value of a derivative at a specific point can be computed several different ways, depending upon how the derivative is computed. The next example illustrates the most common techniques.
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f[x_]= (x2 x + 1)5; f''[1] 30 D[f[x] {x, 2}] /. x 1 , 30 {x, 2}f[x] /. x 1 30 g Derivative[2][f] g[1] 30 f[x_] = x3 g[1] 6
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Here we have de ned a new function, g, as the second derivative of f. If f is changed, g will be the second derivative of the new function. Note the use of here. This is crucial if g is to re ect the change in f. In each of the rst three parts of this example, the second derivative is computed and then x is replaced by 1.
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Differential Calculus
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Mathematica computes derivatives of combinations of functions, sums, differences, products, quotients, and composites by memorizing the various rules. If we do not define the functions, we can see what the rules are.
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Clear[f, g] D[f[x] + g[x], x] f '[x] + g'[x] D[f[x] g[x], x] g[x] f'[x] + f[x] g'[x] D[f[x]/g[x], x]//Together g[x]f'[x] f[x]g'[x] 2 g[x] D[f[g[x]], x] f '[g[x]] g'[x]
Quotient rule. The derivative of a sum is the sum of the derivatives of its terms. This is the familiar product rule.
Chain rule.
We can use Mathematica to investigate some basic theory from a graphical perspective. Rolle s Theorem guarantees, under certain conditions, the existence of a point where the derivative of a function is 0: Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and suppose f(a) = f(b) = 0. Then there exists a number, c, between a and b, such that f ' (c) = 0. In other words, if a smooth (differentiable) function vanishes (has a value of 0) at two distinct locations, its derivative must vanish somewhere in between.
EXAMPLE 13 Show that the function f ( x ) = ( x 3 + 2 x 2 + 15 x + 2)sin x satisfies Rolle s Theorem on the interval
[0, 1] and find the value of c referred to in the theorem. Since f is the product of a polynomial and a trigonometric sine function, f is continuous and differentiable everywhere. f[x_]=(x3 + 2 x2 + 15 x + 2) Sin[o x]; f[0] 0 f[1] 0 FindRoot[f'[c] 0, {c, 0.5}]
0.640241}
We used 0.5 as our initial guess since it is halfway between 0 and 1.
Plot[{f[x], f[.640241]}, {x, 0, 1}]
10 8 6 4 2
Differential Calculus
The Mean Value Theorem is similar to Rolle s Theorem and does not require f to be 0 at each endpoint of the interval: Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists a number, c, between a and b such that f(b) f(a) = f ' (c) (b a). f (b) f (a) If we write the conclusion of the theorem in the form = f '(c), we see that the Mean Value b a Theorem guarantees the existence of a number, c, between a and b, such that the tangent line at (c, f(c)) is parallel to the line segment connecting the endpoints of the curve. Note: Rolle s Theorem and the Mean Value Theorem guarantee the existence of at least one number c. In actuality, there may be several.
EXAMPLE 14 Find the value(s), c, guaranteed by the Mean Value Theorem for the function f ( x ) =
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