course MTH 271 10/19, 1645 ßñ¿óç„ÓxÍ„ú•Ï«ÝŠ³¤z£assignment #013
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01:17:03 2.2.20 der of 4 t^-1 + 1. Explain in detail how you used the rules of differentiation to obtain the derivative of the given function, and give your final result.
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RESPONSE --> s(t)=4t^-1 + 1 s'(t)= -4t^-2 s'(t)= -4/(t^2) Using the Power Rule, nx^(n-1), we are able to find the derivative of the function. n= -1, x=t (-1)(4)(t)^(-1 -1) -4t^-2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:17:20 STUDENT SOLUTION: To solve this using the rules of differentiation, I used the power and constant multiple rules. In dealing with t^-1, I applied the power rule and that gave me derivative -1t^-2. By the constant multiple we multiply this result by the constant 4 to get - 4 t^-2. To deal with 1, I used the constant rule which states that the derivative of a constant is 0. My final result was thus s'(t)=-4t^-2 + 0 = - 4 t^-2. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:35:13 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point?
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RESPONSE --> {3(2 + 'dx)[(4 + 4'dx + 'dx^2) - (2/(2 + 'dx))] - 18 } / 'dx {3(2 + 'dx)[((4 + 4'dx + 'dx^2)(2 + 'dx) -2)/(2+'dx)] -18} / dx (3(8 + 8'dx + 2'dx^2 + 4'dx + 4'dx^2 + 'dx^3 -2) - 18)/ 'dx (3(12'dx + 6'dx^2 + 'dx^3 + 6) - 18) / 'dx ((36'dx + 18'dx^2 + 3'dx^3 + 18) -18) / 'dx (36'dx + 18'dx^2 + 3'dx^3 ) /'dx 'dx(36 + 18'dx + 3'dx^2) / 'dx 36 + 18'dx + 3'dx^2 limit 'dx-->0 36 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:36:44 You could use the product rule with f(x) = 3x and g(x) = x^2 - 2 / x. Since f ' = 3 and g ' = 2 x + 2 / x^2 we have (f g) ' = f ' g + f g ' = 3 (x^2 - 2 / x) + 3x ( 2x + 2 / x^2), which expands to (f g ) ' = 3 x^2 - 6 / x + 6 x^2 + 6 / x. This simplifies to give us just (f g ) ' = 9 x^2. It's easier, though, to just expand the original expression and take the derivative of the result: 3x ( x^2 - 2 / x ) = 3 x^3 - 6. The derivative, using the power-function rule, constant multiple rule and constant rule is thus y ' = 9x^2. At x = 2 we get derivative 9 * ( -2)^2 = 36. Note that (2, 18) is indeed on the graph because 3x ( x^2 - 2/x) evaluated at x = 2 gives us y = 3 * 2 ( 2^2 - 2 / 2) = 6 * 3 = 18. **
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RESPONSE --> Though I did arrive at the correct answer, I obviously took the long route to get there. ------------------------------------------------ Self-critique Rating:ent: 3
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01:38:43 Query 22.2.38 f'(x) for f(x) = (x^2+2x)(x+1) What is f'(x) and how did you get it?
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RESPONSE --> f(x) = (x^2+2x)(x+1) f(x)=(x^3 + x^2 + 2x^2 + 2x) f(x)=x^3 + 3x^2 + 2x f ' (x)=3x^2 + 6x + 2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:38:59 You could use the product rule, which would give you (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' = (2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = 3 x^2 + 6 x + 2. An easier alternative: If you multiply the expressions out you get x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function you get f ' (x) = 3 x^2 + 6 x + 2 . **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:39:46 22.2.66 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative?
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RESPONSE --> C=500 + 7.75x C'=7.75 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:40:25 The terminology means that it costs 7.75 per unit to manufacture the item, and 500 to run the plant or whatever. So if you produce x units it's going to cost 7.75 * x, plus the 500. The cost function is therefore 7.75 x + 500. If you take the derivative of the cost function you are looking at the slope of a graph of cost vs. number produced. The rise between two points of this graph is the difference in cost and the run is the difference in the number produced. When you divide rise by run you are therefore getting the average change in cost, per unit produced, between those two points. That quantity is interpreted as the average cost per additional unit, which is the average variable cost. The derivative is the limiting value of the slope when you let the two graph points get closer and closer together, and so gives the instantaneous rate at which cost increases per additional unit. Note that the fixed cost doesn't influence this rate. Changing the fixed cost can raise or lower the graph but it can't change the slope. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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01:42:48 Why should the derivative of a cost function equal the variable cost?
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RESPONSE --> The derivative of the cost function should equal the variable cost b/c the base cost ($500) is a constant and does not figure into the variable. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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01:42:58 The variable cost is defined as the rate at which the cost changes with repect to the number of units produced. That's the meaning of variable cost. That rate is therefore the derivative of the cost function. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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