lec6 - Lecture 6 18.01 Fall 2006 Lecture 6: Exponential and...

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± ± ± ± Lecture 6 18.01 Fall 2006 Lecture 6: Exponential and Log, Logarithmic Diferentiation, Hyperbolic Functions Taking the derivatives exponentials and logarithms Background We always assume the base, a , is greater than 1. a 0 = 1; a 1 = a ; a 2 = a a ; . . . · a x 1 + x 2 = a x 1 a x 2 ( a x 1 ) x 2 = a x 1 x 2 p q q a = a p (where p and q are integers) r To defne a ±or real numbers r , fll in by continuity. d Today’s main task: ²nd a x dx We can write d a x x x x a = lim - a dx Δ x 0 Δ x We can ±actor out the a x : x x x Δ x Δ x lim a - a = lim a x a - 1 = a x lim a - 1 Δ x 0 Δ x Δ x 0 Δ x Δ x 0 Δ x Let’s call M ( a ) lim a Δ x - 1 Δ x 0 Δ x We don’t yet know what M ( a ) is, but we can say d a x = M ( a ) a x dx Here are two ways to describe M ( a ): d 1. Analytically M ( a ) = a x at x = 0. dx Indeed, M ( a ) = lim a 0+Δ x - a 0 = d a x Δ x 0 Δ x dx x =0 1
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Lecture 6 18.01 Fall 2006 M(a) (slope of a x at x=0) a x Figure 1: Geometric defnition oF M ( a ) x 2. Geometrically, M ( a ) is the slope of the graph y = a at x = 0. The trick to ±guring out what M ( a ) is is to beg the question and de±ne e as the number such that M ( e ) = 1. Now can we be sure there is such a number e ? First notice that as the base a x increases, the graph a gets steeper. Next, we will estimate the slope M ( a ) for a = 2 and a = 4 geometrically. Look at the graph of 2 x in Fig. 2. The secant line from (0 , 1) to (1 , 2) of the graph y = 2 x has slope 1. Therefore, the slope of y = 2 x at x = 0 is less: M (2) < 1 (see Fig. 2).
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lec6 - Lecture 6 18.01 Fall 2006 Lecture 6: Exponential and...

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