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

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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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Presentation transcript:

Definition of the Derivative Using Average Rate (Page 129 - 133 and 160 in the book) f(a+h) Slope of the line = Average Rate of Change = f(a+h) – f(a) h f(a+h) – f(a) f(a) h 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 a+h h a a+h h a a h

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

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

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

If f(x) = x2 - 8x + 9, then f ' (x) = 2x - 8 Example: Use the definition of the derivative to obtain the following result: If f(x) = x2 - 8x + 9, then f ' (x) = 2x - 8 f(x+h) – f(x) h f ' (x) = lim h 0 Solution: Using the definition f (x + h) = (x + h)2 - 8(x + h) + 9 = (x2 + 2xh + h2 - 8x -8h + 9) f (x + h) – f (x) h f ' (x) = lim h 0 = (x2 + 2xh + h2 - 8x - 8h + 9) – ( x2 - 8x + 9) h lim h 0 = (2xh + h2 - 8h) h lim h 0 = h (2x + h - 8) h lim h 0 = (2x + h - 8) lim h 0 = 2x - 8