So with the equation MU=ln(C1)+ln(160-1.1C1) the derivative is

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In summary: Dividing by 2.2 gives 2.2C1= 160. Therefore, in summary, the steps to solve for the derivative being equal to 0 are: 1. Add 1.1/(160- 1.1C1) to both sides. 2. Multiply both sides by C1 and 160- 1.1C1. 3. Add 1.1C1 to both sides. 4. Divide by 2.2 to get the solution, 2.2C1= 160.
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surferbarney0729
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So with the equation MU=ln(C1)+ln(160-1.1C1)

the derivative is 1/C1 - 1.1/(160-1.1C1)

How does one solve and identify the derivative instead being seen and written as

2.2C1 = 160

What are the steps taken to get to the latter derivative. the second format was much easier to solve the final equation, but I am not sure how to 2.2C1 = 160 was found

Any help?
 
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The equation you want is from 1/C1 - 1.1/(160-1.1C1) = 0
 
  • #3


swoodward said:
So with the equation MU=ln(C1)+ln(160-1.1C1)

the derivative is 1/C1 - 1.1/(160-1.1C1)

How does one solve and identify the derivative instead being seen and written as

2.2C1 = 160

What are the steps taken to get to the latter derivative. the second format was much easier to solve the final equation, but I am not sure how to 2.2C1 = 160 was found

Any help?
The "second format" cannot be derived from the first without additional information. I presume that the actual problem was to find where the derivative is equal to 0: 1/C1- 1.1/(160- 1.1C1)= 0. Add 1.1/(160- 1.1C1) to both sides to get
1/C1= 1.1/(160- 1.1C1). Now multiply both sides by C1 and 160- 1.1C1 to get rid of the fraction: 160- 1.1C1= 1.1C1. Adding 1.1C1 to both sides gives 160= 2.2C1.
 

Related to So with the equation MU=ln(C1)+ln(160-1.1C1) the derivative is

1. What is the equation MU=ln(C1)+ln(160-1.1C1)?

The equation MU=ln(C1)+ln(160-1.1C1) represents the marginal utility (MU) of a certain product or service, which is calculated by taking the natural logarithm of the quantity (C1) consumed, plus the natural logarithm of the remaining quantity (160-1.1C1) available.

2. What does the derivative of this equation represent?

The derivative of MU=ln(C1)+ln(160-1.1C1) represents the rate of change of the marginal utility with respect to the quantity consumed (C1). In other words, it shows how much the marginal utility will change for a small change in the quantity consumed.

3. How is the derivative calculated for this equation?

The derivative of MU=ln(C1)+ln(160-1.1C1) can be calculated using the logarithmic differentiation rule, which states that the derivative of ln(x) is equal to 1/x. By applying this rule to both terms in the equation and using the chain rule, the derivative can be found.

4. Why is the derivative of this equation important?

The derivative of MU=ln(C1)+ln(160-1.1C1) is important because it allows us to analyze the relationship between the quantity consumed and the marginal utility. It helps us understand how changes in the quantity consumed will affect the marginal utility, and can be used to optimize the consumption of a product or service.

5. Are there any limitations to using this equation and its derivative?

Like any mathematical model, there are limitations to using the equation MU=ln(C1)+ln(160-1.1C1) and its derivative. It assumes a linear relationship between quantity consumed and marginal utility, which may not always hold true in real-world scenarios. Additionally, it may not account for other factors that may influence marginal utility, such as individual preferences and external factors.

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