alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# terms with higher powers of x a l 3 limit step now

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Unformatted text preview: -- f(a) b -- a INSTANT ANTANEOUS RATE 14. INSTANTANEOUS RATE OF CHANGE of xn Let us consider how to derive the general expression for I N S TA N TA N E O U S R AT E O F C H A N G E of the general polynomial : a n x n + a n -- 1 x n -- 1 + . . . + a 1 x + a 0 Obser ve that the ter ms of this function are of the for m a n x n for n = 0, 1, 2, 3, . . . To DIFFERENTIATE the general polynomial we should be able to find the INSTANTANEOUS RATE OF CHANGE of the general variable term x n . We can think of x n a s the function x n. In other words, we should be able to DIFFERENTIATE the function f(x) = x n. As usual we will first find the INSTANTANEOUS RATE OF CHANGE of the function f(x) = xn at par ticular point a . Then we will find the general expression of the particular par any INSTANTANEOUS RATE OF CHANGE f(x) = xn at any point x. an We note that : Value x=a f(x) = an We can differentiate f(x) = xn at x = a in two ways : 1. The Vedic way - without the use of δ x . 2. The Western way - using the infinitesimal δ x . 65 wa Vedic w ay : f ’ (a) = (a) Limit x→a AVERAGE RATE OF CHANGE close to a . = x n -- a n x -- a CHANGE in function CHANGE in variable So : f(x) − f(a) xn − an = x−a x−a Properly speaking we have to compute both : L imit f(x) − f(a) L imit f(x) − f(a) and x → a+ x−a x−a But in the Vedic way the Algebraic notation is lacking. For n = 1 and n = 2 the reader may wish to review example 4 (page 28) and example 5 (page 29). x → a-- For n = 3 : Limit x 3--a 3 = Limit ( x--a) (x 2 +xa+a 2 ) Limit = x → a (x 2 +xa+a 2 ) x → a x--a x→a (x--a) Substitute for x = a to get this Limit = 3a 2 Let us do this in three steps for general n. VERAGE step 1. AVERAGE step : f(x) − f(a) xn − an = x−a x−a 2. TENDS TO step : x n--a n = (x − a)(x n--1 + x n--2 a + x n--3 a 2 + . . . + xa n-- 2 + a n-- 1 ) x--a x--a = (x n--1 + x n--2 a + x n--3 a 2 + . . . + xa n-- 2 + a n-- 1 ) We may simplify because (x − a) ≠ 0 . 3. LIMIT step : Limit (x n--1 + x n--2 a + x n--3 a 2 + . . . + xa n-- 2...
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