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4.2 logs.pdf

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Dan Dolderman

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Calc 1 Tues 10/7/08 4.2 - Derivatives of Log functions Well, if we’re going to ▯gure out what the derivative of ln(x) is, we might as well get cracking! d ln(x + h) ▯ ln(x) [ln(x)] = lim dx h!0 h 1 = lim [ln(x + h) ▯ ln(x)] h!0 h ▯ ▯ 1 x + h = lim ln h!0 h ▯ x ▯ 1 h = lim ln 1 + h!0 h x h Okay, now let’s do a substitution. Letxv =Notice that if we think of x as ▯xed, then when we take h ! 0, then v ! 0. This means 1 = lim ln(1 + v) v!0vx 1 1 = lim ln(1 + v) xv!0 v 1 1=v = lim ln(1 + v) xv!0 Oh, but we’ve seen that limit before! ...Or at least a variation of it. We’ve seen ▯ ▯ 1 x lim 1 + x!1 x which is exactly the same thing. What’s that limit? Why, it’s e! So, our limit is just 1 1 lim ln(e) = x v!0 x All right, so[ln(x)] =1, so long as x > 0. Also, for any base b, dx x d [log (x)] = 1 dx b xln(b) Example: ▯nd d [ln(2x + 3)]. dx Solution: This is just chain rule, so derivative of the outside woul, and then multiply 2x+3 that by the derivative of the inside, which is 2. So, the derivative will be 2 2x + 3 Since it’s so much easier taking the derivatives of sums and di▯erences rather than products and quotients (because you have to use those rules!), a lot of times, it’s much better to use log properties to split up the functions into sums and then take the derivative. 1 Example: ▯ ▯ d x sin(x) d ▯ ▯ ln 2 = ln(x ) + ln(sin(x)) ▯ ln(tan(x + 1)) dx tan(x + 1) dx 2 2 2 = 3x + cos(x)▯ 2xsec (x + 1) x3 sin(x) tan(x + 1) 2 2 3 2xsec (x + 1) = x + cot(x) ▯ tan(x + 1) Oh, and by the way, you can extend the derivative of log functions to d 1 lnjxj = ; x 6= 0 dx x This will become more important when we do integrals later on. But for now, let’s move on to
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