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Definition of the Derivative Using Average Rate () a a+h f(a) Slope of the line = h f(a+h) Average Rate of Change = f(a+h) – f(a) h f(a+h) – f(a) h

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a+h h a h a a h Now, Watch what happens when: Point a is fixed and the size of the interval h shrinks a+h h a

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As h shrinks and approaches zero (but not = 0), Slope of the Tangent line = f(a+h) – f(a) h h f(a+h) f(a) a+ha Slope of the line = Average Rate of Change = f(a+h) – f(a) h the line becomes a Tangent Line As h approaches zero

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f(a) a lim: Limit, as h approaches zero f(a+h) – f(a) h = lim h 0 Slope of the Tangent line As h approaches zero, or: f(a+h) – f(a) h = The slope of the Tangent Line at a is the Derivative, f ' (a) f(a+h) – f(a) h lim h 0 f ' (a) =

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Example: Use the definition of the derivative to obtain the following result: If f(x) = -2x + 3, then f ' (x) = -2 Solution: Using the definition f (x + h) = -2(x + h) + 3= (-2x - 2h + 3) = -2 f(x+h) – f(x) h f ' (x) = lim h 0 = (-2h) h lim h 0 = (-2x - 2h + 3) – ( -2x + 3) h lim h 0 f (x + h) – f (x) h f ' (x) = lim h 0

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Example: Use the definition of the derivative to obtain the following result: If f(x) = x 2 - 8x + 9, then f ' (x) = 2x - 8 f (x + h) = (x + h) 2 - 8(x + h) + 9= (x 2 + 2xh + h 2 - 8x -8h + 9) = 2x - 8 f(x+h) – f(x) h f ' (x) = lim h 0 = (2x + h - 8) lim h 0 = h (2x + h - 8) h lim h 0 = (2xh + h 2 - 8h) h lim h 0 = (x 2 + 2xh + h 2 - 8x - 8h + 9) – ( x 2 - 8x + 9) h lim h 0 f (x + h) – f (x) h f ' (x) = lim h 0 Solution: Using the definition

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Slope Fields. Quiz 1) Find the average value of the velocity function on the given interval: [ 3, 6 ] 2) Find the derivative of 3) 4) 5)

Slope Fields. Quiz 1) Find the average value of the velocity function on the given interval: [ 3, 6 ] 2) Find the derivative of 3) 4) 5)

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