1,2 Class Notes - 3.4 and 4.1. Defining and Computing...

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3.4 and 4.1. Defining and Computing Derivatives The derivative of f is the instantaneous rate of change 0 () ( ) '( ) lim h fx h fx fx h +− = Derivative is the slope of the tangent line Slopes -23, -14, -7.52, -6 70 75 80 85 90 95 100 3.9 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5 Constant functions: f(x)=5 0 55 lim 0 h h == The curve is flat so the slope is 0 Linear functions: f(x) = 3x+2 3( ) 3 3 3 xh x h hh Constant slope =3, so derivative = 3 f(x) = x 2 22 2 2 2 0 2 2 2 2 lim 2 2 h ab a a bb x x xh x x += + + + + + + f(x) = x 3 33 2 23 2 0 3 3 lim 3 3 3 h x x h x h h x xx h h h h h x + + + =++ ++ =
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Power functions Function Derivative x1 x 2 2x x 3 3x 2 x p px p-1 Constant Multiples Rule 1: If f(x) = x p then f (x) = px p-1 Rule 2: If c is a number, (cf) = cf . f(x) = 4x 3 f (x) = 4(3x 2 ) = 12x 2 f(x) = -7x 5 f (x) = (-7)(5x 4 ) = -35x 4 Derivative of Sums Rule 1: (x p ) = px p-1 Rule 2: (cf) = cf Rule 3. (f+g) = f + g f(x) = 10 + 18x + 9x 2 –2x 3 f (x) = 0 + 18(1) + 9(2x) – 2(3x 2 ) f (4) = 18 + 72 – 96 = -6 Practice Problem f(x) = x 4 –3x 3 +7x 2 + 11 Find f (x) and f (1) Weight of Bighorn sheep on Ram Mountain in Alberta, Canada: M(t) = 27.5 + 0.3 t – 0.001 t 2 M(t) is measured in kilograms t = number of days since May 25. Find the instantaneous rate of change of weight for a sheep at t = 100 days. M(t) = 27.5 + 0.3 t – 0.001 t
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1,2 Class Notes - 3.4 and 4.1. Defining and Computing...

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