#$&* course Mth 271 11/26 430 013. `query 13*********************************************
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Given Solution: `a STUDENT SOLUTION: To solve this using the rules of differentiation, I used the power and constant multiple rules. In dealing with t^-1, I applied the power rule and that gave me derivative -1t^-2. By the constant multiple we multiply this result by the constant 4 to get - 4 t^-2. To deal with 1, I used the constant rule which states that the derivative of a constant is 0. My final result was thus s'(t)=-4t^-2 + 0 = - 4 t^-2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q 22.2.30 der of 3x(x^2-2/x) at (2,18) What is the derivative of the function at the given point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My first step is to continue with the original function and expand before taking the derivative. 3x(x^2-2/x) 3x^3-6x/x 3(3)x^2 - 6 9x^2 is the derivative of the original function. At x+2 we have 9(2)^2= 9(4)= 36 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: I multiplied to two sets together to get x^3+x^2+ 2x^2+ 2x. Added like terms for x^3 + 3x^2 + 2x. using the power rule we get 3x^2 + 6x + 2= f'(x) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could use the product rule, which would give you (x^2 + 2x) ' ( x + 1) + (x^2 + 2x) ( x + 1) ' = (2x + 2) ( x + 1) + (x^2 + 2 x ) ( 1) = 2 x^2 + 4 x + 2 + x^2 + 2 x = 3 x^2 + 6 x + 2. An easier alternative: If you multiply the expressions out you get x^3+3x^2+2x. Then applying the constant multiple rule and the simple power rule to the function you get f ' (x) = 3 x^2 + 6 x + 2 . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q 22.2.66 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If it costs 7.75 per unit and a cost of 500 added regardless of how many units the cost function would simply be 7.75x + 500. x being any variable of choice. f'(x)= 7.75. 500 being a constant has a derivative of 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qWhy should the derivative of a cost function equal the variable cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative gives you the rate at which it changes, which is exactly what the variable cost is. Variable cost being the amount, cost, per unit. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!