alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Average ste p constant x p2 p1 fx2 fx1 c

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Unformatted text preview: VALUED, CONTINUOUS and DIFFERENTIABLE): f ’ (x) = L imit → δx 0 f (x + δ x) -- f (x) δx Instead of using the symbols δ x we will use the letter h. So, by our understanding, we can define the DERIVATIVE of a function to be: f ’ (x) = f (x + h ) -- f (x) h Now, let us use our definition of the FIRST DERIVATIVE to prove that : L imit h→ 0 dxn n x n -- 1 dx = 68 Proof Proof : h f( x+h ) = ( x+ h )n = x n ( 1 + -- ) x n f(x) = x VERAGE step 1. AVERAGE ste p : f (x + h ) -- f (x) = h 1 h -- [ x n ( 1 + -- ) n -- x n ] x h xn h --- [ ( 1 + -- ) n -- 1] x h 2. TENDS TO step : expanding by the Binomial Theorem : = n h n( n -- 1 ) -- 2 + . . . terms -x [ ( 1 + n -- + -(h) x x h 2! with higher powers of h ) -- 1 ] The +1 and -- 1 will cancel out. n h n( n -- 1 ) -- 2 + . . . terms with higher powers of h ] -= -x [ ( n -- + (h) x x h 2! / Since h → 0 but h = 0 we may cancel h in the numerators and denominator. = n n( n -- 1) --- + . . . terms with higher powers of h ] h x n [ ( -- + ( x2 ) x 2! 3. LIMIT step : now we can take the LIMIT as h → 0 . h L imit h = 0 , L imit --- = 0 and L imit (terms with higher powers of h) = 0. h→ 0 h → 0 x2 h→ 0 = We are left with : n x n -- = n x n -- 1 x We denote this INSTANTANEOUS RATE OF CHANGE as: d --- xn = n x n--1 dx This is also called the FIRST DERIVATIVE of x n. 69 Most of the confusion in Calculus arises out of the fear of dividing by zero. Limit δ x TENDS TO zero is zero. Therefore, it appears in L imit δ y to be δx → 0 δ x dividing by zero. We must always keep in mind that δx ia attached to some par ticular instant particular par eneral a or general instant x . As δx → 0 we approach that INSTANT computing as we go along the AVERAGE RATE OF CHANGE δ y . δx When we take L imit of this AVERAGE RATE OF CHANGE δ y we get the δx δx → 0 d y at the par ticular instant a or particular INSTANTANEOUS RATE OF CHANGE par dx general instant x . Exer ercise Exercise : Prove the INSTANTANEOUS RATE OF CHANGE of a n x n = n a n x n − 1 ?...
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