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Chapter 3 sections 3.3 and 3.4 What is differentiation and what are the methods of differentiation?
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Section 3.3 Rules of Differentiation day 1 What is the power rule for derivatives?
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Power Rules Find the derivative of each: f(x)f(x) = x 2 f(x) = x 3 Do you see a pattern?
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explore Does the pattern hold now? f(x) = 2xf(x) = 3x 2 What modification has to be made?h
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Derivative of a Constant Find the derivative f(x) = 2f(x) = 3 Would it help to think of these like this? f(x) = 2x 0 f(x) = 3x 0 Find the derivative: consider it in graphical form. What is the Slope of the line represented by A constant?
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Sum & Difference Differentiation (differentiation of a polynomial) Find the derivative: f(x) = 3x 2 + 2x (hint: The limit of a sum = the sum of the limits— take the derivative of each part) f(x) = 3x 2 + 2x
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Product Rule Find the derivative of the product If g(x) = x + 2 and h(x) = 3x 2 + 1 f(x) = (x + 2)(3x 2 + 1) f(x) = 3x 3 + 6x 2 + x + 2 f‘(x) = 9x 2 + 12 x + 1 Do you really want to multiply these to get the solution? If r(x) = x 3 + 2x 2 + 3x + 5 And s(x) = 4x 5 – 2x 3 +3x 2 – 6x + 1 Find g’ and h’ g‘ (x) = 1 and h’(x) = 6x TRY: f’(x) = g(x)h’(x) + h(x)g’(x) f‘(x) = (x+2)(6x) + (3x 2 + 1)(1) = 6x 2 + 12x + 3x 2 + 1 = 9x 2 + 12 x + 1 Rule=1 st d(2 nd ) + 2 nd d(1 st )
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Quotient Rule:
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Consider
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f(x) = 3x 3 – 4x 2 + 2x – 5 Find f”(x)
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Homework Page(s): 120 2 – 20b
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Section 3.3 day 2 What is the power rule for derivatives?
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Homework Page(s): 120 24, 26, 28, 30
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Section 3.3 day 3 What is the power rule for derivatives?
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IN CLASS Page(s): 121 33, 34, 35
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Section 3.4 Velocity & rates of change day 1 How is the rate of change related to the derivative?
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Rate of Change Rate of Change is equivalent to instantaneous rate of change Therefore the derivative of a function
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Example 1 Find the rate of change of the area of a circle with respect to its radius A= πr 2 A’ = 2πr When r = 5 A’ = 10π
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Relationship between distance velocity and acceleration Motion Distance (s) = f(x) Velocity (v) = f’(x) Acceleration (a) = f”(x)
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Example A projectile is hurled upwards and follows a path defined by s=6 + 64 t + 16t 2 Find the max height of the object S = – 16t 2 + 64 t + 6 V = -32t + 64 A = -32 What do we know about the graph? 2 nd power indicates parabolic flight Neg on the leading coefficient indicates a max pt. (max and mins occur when the tangent is horizontal, so the slope of the tangent is 0)
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Example S = – 16t 2 + 64 t + 6 V = -32t + 64 A = -32 0 = -32t + 64 -64 = -32t 2 = t The max occurs when t = 2 at S = – 16(2) 2 + 64(2) – 6 = 72 ft.
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Example S = – 16t 2 + 64 t + 6 V = -32t + 64 A = -32 How fast is it traveling when it is 54’ high? 54 = – 16t 2 + 64 t + 6 0 = – 16t 2 + 64 t – 48 t = 3 or 1 V(1) = 32 ft/sec v(3) = -32 ft/sec
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Example S = – 16t 2 + 64 t + 6 V = -32t + 64 A = -32 What is the acceleration? A = -32 ft/sec 2 (it starts to slow down as soon as it leaves the ground)
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Example S = – 16t 2 + 64 t + 6 V = -32t + 64 A = -32 When does it hit the ground? 0 = – 16t 2 + 64 t + 6 Use the quad formula T = 4.09 sec or -.09 sec What does this tell us? The object did not start on the ground Where did it start? S = – 16(0) 2 + 64(0) + 6 = 6’
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Uses of derivatives Economics Cost equations =c(x) Marginal cost = c’(x) Revenue = r(x) Marginal revenue = r’(x)
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Homework Page(s): 129 1, 3, 5, 9, 13
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Section 3.4 day 2 How is the rate of change related to the derivative?
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Homework Page(s): Review
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