course Mth 271
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11:37:10 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 --> 4(-1) t^-1-1 ....and the one goes away because there isnt a variable attached to it -4t^-2
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11:37:52 ** 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 --> I didnt explain it so well but I got the correct answer
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11:40:34 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 --> 3x(x^2)= 3x^3 - (2/x * 3x)= 3x^3 - 6 f(2)= 3(2)^3-6 =18 ......when x= 2 then y= 18
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11:41:43 ** 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 --> I got all the way to the point where I needed to find the deriv. and just forgot the whole step...I know how this works I am just not thinking
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11:43:56 Query 22.2.40 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 --> x^2*x=x^3 x^2*1= x^2 2x*x= 2x^2 2x*1=2x x^3+ x^2+2x^2+2x '=3x^2+ 2x+ 4x+2 = 3x^2+6x+2
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11:44:14 ** 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 --> This is what I got
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11:45:21 22.2.56 vbl cost 7.75/unit; fixed cost 500 What is the cost function, and what is its derivative?
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RESPONSE --> cost function= 7.75x+500 cf'= 7.75 =variable
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11:45:39 ** 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 --> I understand this
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11:46:32 Why should the derivative of a cost function equal the variable cost?
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RESPONSE --> the rate= variable cost and the ' = cost function
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11:46:45 ** 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 --> I understand this ok
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