barcode print in asp net THE DERIVATIVE in .NET

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THE DERIVATIVE
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RULE 3 (Product Rule).
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Dx ( f (x) g(x)) = f (x) Dx g(x) + g(x) Dx f (x)
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For a proof, see Problem 13.1. EXAMPLES
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(a) Dx (x 4 ) = Dx (x 3 x) algebra ua ub = ua+b = x 3 Dx (x) + x Dx (x 3 ) = x 3 (1) + x(3x 2 ) = x 3 + 3x 3 = 4x 3 (b) Dx (x 5 ) = Dx (x 4 x) = x 4 Dx (x) + x Dx (x 4 ) = x 4 (1) + x(4x 3 ) = x 4 + 4x 4 = 5x 4 (c) Dx ((x 3 + x)(x 2 x + 2)) = (x 3 + x) Dx (x 2 x + 2) + (x 2 x + 2) Dx (x 3 + x) = (x 3 + x)(2x 1) + (x 2 x + 2)(3x 2 + 1) [by the product rule] [by example (a)] and ua = ua b ub
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[by the product rule]
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The reader may have noticed a pattern in the derivatives of the powers of x: Dx (x) = 1 = 1 x 0 Dx (x 2 ) = 2x Dx (x 3 ) = 3x 2 Dx (x 4 ) = 4x 3 Dx (x 5 ) = 5x 4
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This pattern does in fact hold for all powers of x. RULE 4. Dx (x n ) = nx n 1 where n is any positive integer. For a proof, see Problem 12.2. EXAMPLES
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(a) Dx (x 9 ) = 9x 8 (b) Dx (5x 11 ) = 5 Dx (x 11 ) = 5(11x 10 ) = 55x 10
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Using Rules 1, 2, and 4, we have an easy method for differentiating any polynomial. EXAMPLE
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Dx 1 3 3 x 4x 2 + 2x 5 2 = Dx = 3 3 1 x Dx (4x 2 ) + Dx (2x) Dx 5 2 [by Rule 1] [by Rule 2 and Corollary 12.2] [by Rule 4]
3 Dx (x 3 ) 4 Dx (x 2 ) + 2 Dx (x) 0 5 3 = (3x 2 ) 4 (2x) + 2 (1) 5 9 = x 2 8x + 2 5
THE DERIVATIVE
[CHAP. 12
More concisely, we have: RULE 5. To differentiate a polynomial, change each nonconstant term ak x k to kak x k 1 and drop the constant term (if any). EXAMPLES
(a) Dx (8x 5 2x 4 + 3x 2 + 5x + 7) = 40x 4 8x 3 + 6x + 5 (b) Dx 3x 7 + 5 4 2 2x x + 9x 3 8 = 21x 6 + 5 2x 4 x + 9 3
Solved Problems
12.1 Prove: (a) Rule 1(i, ii); (b) Rule 2. Assume that Dx f (x) and Dx g(x) are de ned.
[ f (x + h) g(x + h)] [ f (x) g(x)] h [ f (x + h) f (x)] [g(x + h) g(x)] h f (x + h) f (x) g(x + h) g(x) h h f (x + h) f (x) g(x + h) g(x) lim h h h 0 [by Section 8.2, Property V]
(a) Dx ( f (x) g(x)) = lim = lim = lim
= lim
= Dx f (x) Dx g(x) (b) Dx (c f (x)) = lim c f (x + h) c f (x) c[ f (x + h) f (x)] = lim h h h 0 f (x + h) f (x) = c lim [by Section 8.2, Property III] h h 0 = c Dx f (x)
12.2 Prove Rule 4, Dx (x n ) = nx n 1 , for any positive integer n.
We already know that Rule 4 holds when n = 1, Dx (x 1 ) = Dx (x) = 1 = 1 x 0 (Remember that x 0 = 1.) We can prove the rule by mathematical induction. This involves showing that if the rule holds for any particular positive integer k, then the rule also must hold for the next integer k + 1. Since we know that the rule holds for n = 1, it would then follow that it holds for all positive integers. Assume, then, that Dx (x k ) = kx k 1 . We have Dx (x k+1 ) = Dx (x k x) = x k Dx (x) + x Dx (x k ) = x k 1 + x(kx k 1 ) = x k + kx k = (1 + k)x k = (k + 1)x (k+1) 1 and the proof by induction is complete. [since x k+1 = x k x 1 = x k x] [by the product rule] [by the asumption that Dx (x k ) = kx k 1 ] [since x x k 1 = x 1 x k 1 = x k ]
CHAP. 12]
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