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3.2 derivatives.pdf

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Dan Dolderman

Calc 1 Mon 9/22/08 3.2 - Derivatives! 0 Derivative: The slope of the tangent line. It’s all that stu▯ we did last week. Denoted f (x) (pronounced f prime of x), it’s simply 0 f(x + h) ▯ f(x) f (x) = lim h!0 h and is called the derivative with respect to x. 0 The domain of f (x) is all the x in the domain of f(x) AND where the above limit exists. 2 Example: Find the derivative with respect to x of f(x) = 3x ▯ 2. Solution: f (x) = limf(x + h) ▯ f(x) h!0 h 2 2 [3(x + h) ▯ 2] ▯ [3x ▯ 2] = lim h!0 h [3(x + 2xh + h ▯ 2] ▯ 3x + 2 = lim h!0 h 3x + 6xh + 3h ▯ 2 ▯ 3x + 2 = lim h!0 h 2 = lim6xh + 3h h!0 h = limh(6x + 3h) h!0 h = h!0m 6x + 3h = 6x Keep in mind that the derivative is a function, so you should be getting out something with x in it. Only when we want to know what the derivative is at a speci▯c point (like, say, x = 3) do we only get a number out. The derivative is the function that when you plug in a value for x, it spits out the slope of f(x) at that value for x. Compare the graphs of f(x) = 3x ▯ 2 and f (x) = 6x. 1 Example: Let’s take a look at what the derivative of some line y = mx + b might be. Solution: 0 f(x + h) ▯ f(x) f (x) = lim h!0 h [m(x + h) + b] ▯ [mx + b] = lim h!0 h mx + mh + b ▯ mx ▯ b = lim h!0 h mh = lim h!0 h = lim m h!0 = m So, the slope of the tangent line of a line is just the slope of the line itself! Well, that actually kind of makes sense if we look at the graph of any ol’ line! Now here’s a shocker: for f(t) a position function of a particle in rectilinear motion, the derivative 0 f (t) is the instantaneous velocity function and is (what else but) 0 f(t + h) ▯ f(t) v(t) = f (t) = lim h!0 h Technically, it’s the instantaneous velocity function, but just simply call it the velocity function. If we want average velocity, we’ll ask for the average velocity. Example: Say a particle is moving by the position function f(t) = 4t + t, where f(t) is measured in meters and t in seconds. Find the velocity function. Solution: [4(t + h) + (t + h)] ▯ [4t + t] f (t) = lim h!0 h [4(t + 2th + h ) + t + h] ▯ 4t ▯ t = lim h!0 h 2 2 2 = lim 4t + 8th + 4h + t + h ▯ 4t ▯ t h!0 h 2 8th + 4h + h = h!0 h h(8t + 4h + 1) = h!0 h = lim 8t + 4h + 1 h!0 = 8t + 1 Okay, so here I am talking about all this derivative stu▯, but I neglected to mention the fact that the derivative doesn’t have to always exist. Remember when I ▯rst de▯ned it, I mentioned that the domain of the derivative was the x’s in the
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