#$&* course Mth 271 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a STUDENT SOLUTION: To solve this using the rules of differentiation, I used the power and constant multiple rules. In dealing with t^-1, I applied the power rule and that gave me derivative -1t^-2. By the constant multiple we multiply this result by the constant 4 to get - 4 t^-2. To deal with 1, I used the constant rule which states that the derivative of a constant is 0. My final result was thus s'(t)=-4t^-2 + 0 = - 4 t^-2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I had to review the instructors solution. 3x ( x^2 - 2 / x ) = 3 x^3 - 6. From the derivative, using the power-function rule, constant multiple rule and constant rule is thus y ' = 9x^2. At x = 2 we get derivative 9 * ( -2)^2 = 36. Note that (2, 18) is indeed on the graph because 3x ( x^2 - 2/x) evaluated at x = 2 gives us y = 3 * 2 ( 2^2 - 2 / 2) = 6 * 3 = 18. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 22.2.38 f'(x) for f(x) = (x^2+2x)(x+1) What is f'(x) and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function = 3 x^2 + 6 x + 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could use the product rule, which would give you (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' = (2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = 3 x^2 + 6 x + 2. An easier alternative: If you multiply the expressions out you get x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function you get f ' (x) = 3 x^2 + 6 x + 2 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 22.2.66 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7.75x + 500 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. The derivative of the cost function is then easily found to be • dC / dx = 7.75. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy should the derivative of a cost function equal the variable cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The variable cost is defined as the rate at which the cost changes with respect to the number of units produced. The rate would be the y’ of the cost f(x). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The variable cost is defined as the rate at which the cost changes with respect to the number of units produced. That's the meaning of variable cost. That rate is therefore the derivative of the cost function. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 271 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I had to review the instructors solution. 3x ( x^2 - 2 / x ) = 3 x^3 - 6. From 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. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could use the product rule with f(x) = 3x and g(x) = x^2 - 2 / x. Since f ' = 3 and g ' = 2 x + 2 / x^2 we have (f g) ' = f ' g + f g ' = 3 (x^2 - 2 / x) + 3x ( 2x + 2 / x^2), which expands to (f g ) ' = 3 x^2 - 6 / x + 6 x^2 + 6 / x. This simplifies to give us just (f g ) ' = 9 x^2. It's easier, though, to just expand the original expression and take the derivative of the result: 3x ( x^2 - 2 / x ) = 3 x^3 - 6. The derivative, using the power-function rule, constant multiple rule and constant rule is thus y ' = 9x^2. At x = 2 we get derivative 9 * ( -2)^2 = 36. Note that (2, 18) is indeed on the graph because 3x ( x^2 - 2/x) evaluated at x = 2 gives us y = 3 * 2 ( 2^2 - 2 / 2) = 6 * 3 = 18. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 22.2.38 f'(x) for f(x) = (x^2+2x)(x+1) What is f'(x) and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function = 3 x^2 + 6 x + 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 22.2.66 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7.75x + 500 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The terminology means that it costs 7.75 per unit to manufacture the item, and 500 to run the plant or whatever. So if you produce x units it's going to cost 7.75 * x, plus the 500. The cost function is therefore 7.75 x + 500. The derivative of the cost function is then easily found to be • dC / dx = 7.75. If you take the derivative of the cost function you are looking at the slope of a graph of cost vs. number produced. The rise between two points of this graph is the difference in cost and the run is the difference in the number produced. When you divide rise by run you are therefore getting the average change in cost, per unit produced, between those two points. That quantity is interpreted as the average cost per additional unit, which is the average variable cost. The derivative is the limiting value of the slope when you let the two graph points get closer and closer together, and so gives the instantaneous rate at which cost increases per additional unit. Note that the fixed cost doesn't influence this rate. Changing the fixed cost can raise or lower the graph but it can't change the slope. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhy should the derivative of a cost function equal the variable cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The variable cost is defined as the rate at which the cost changes with respect to the number of units produced. The rate would be the y’ of the cost f(x). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The variable cost is defined as the rate at which the cost changes with respect to the number of units produced. That's the meaning of variable cost. That rate is therefore the derivative of the cost function. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!