Assignment 13 query

course Mth 271

013. `query 13*********************************************

Question: `q2.2.20 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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Your solution:

y = 4t^-1 + 1

y’ = 4(-1)t^(-1 -1) + 1^0

y’ = -4t^-2

confidence rating:

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I feel very confident about this problem.

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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. **

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Self-critique Rating:

I used the power rule to find the answer to this problem.

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Question: `q 22.2.30 derivative of 3x(x^2-2/x) at (2,18)

What is the derivative of the function at the given point?

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Your solution:

y = 3x(x^2 – 2/x)

y’ = f’g + fg’

y’ = 3(x^2 – 2/x) + 3x(2x -2/x^2)

y’ = 3x^2 – 6/x + 6x^2 – 6/x

y’ = 9x^2

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel very confident about this problem.

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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. **

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Self-critique Rating:

I used the product rule.

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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?

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Your solution:

y = (x^2 + 2x)(x + 1)

y’ = f’g + fg’

y’ = (2x + 2)(x + 1) + (1)(x^2 + 2x)

y’ = 2x^2 + 2x + 2x + 2 + x^2 + 2x

y’ = 3x^2 + 6x + 2

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel very confident about this problem.

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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 . **

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Self-critique Rating:

I used the product rule.

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Question: `q 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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Your solution:

The cost function would be: 7.75x + 500

The derivative would be: 7.75

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel very confident about this problem.

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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. **

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Self-critique Rating:

I used logic to answer this problem.

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Question: `qWhy should the derivative of a cost function equal the variable cost?

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Your solution:

Because the variable cost is the rate at which it costs to manufacture the product. Because it is a rate, it is the derivative.

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel very confident about this problem.

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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. **

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Self-critique Rating:

I used logic to answer this problem.

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&#Very good responses. Let me know if you have questions. &#