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Derivative of a Utility Function

  • Thread starter Dakarai
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Homework Statement


What is the MRS of the quasilinear utility function U(q1, q2) = u(q1) + q2 ?

Homework Equations


MRS = - dU1/dU2

The Attempt at a Solution


[/B]dU2 is 1 but I am unsure how to approach taking the derivative of u(q1). I have tried the answer as -dU and -dU * dq1, but neither of those are the correct answer.
 

Answers and Replies

  • #2
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Homework Statement


What is the MRS of the quasilinear utility function U(q1, q2) = u(q1) + q2 ?

Homework Equations


MRS = - dU1/dU2
What is MRS? I'm guessing that this question is from an economics course.

In the problem statement above, you have functions U and u, and variables q1 and q2. What is meant by dU1 and dU2? The notation here is very terse. Is MRS the quotient of the two partials of U? I.e.,
$$- \frac{\frac{\partial U}{\partial q_1}}{\frac{\partial U}{\partial q_2}}?$$
Dakarai said:

The Attempt at a Solution


[/B]dU2 is 1 but I am unsure how to approach taking the derivative of u(q1). I have tried the answer as -dU and -dU * dq1, but neither of those are the correct answer.
 
  • #3
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What is MRS? I'm guessing that this question is from an economics course.

In the problem statement above, you have functions U and u, and variables q1 and q2. What is meant by dU1 and dU2? The notation here is very terse. Is MRS the quotient of the two partials of U? I.e.,
$$- \frac{\frac{\partial U}{\partial q_1}}{\frac{\partial U}{\partial q_2}}?$$
Sorry I wasn't very clear! (I'm not familiar with using that fancy notation.) MRS stands for Marginal Rate of Substitution, the rate that one consumer is willing to give up in exchange for another while keeping the same utility (U). MRS is the quotient of the two partials as you wrote: $$- \frac{\frac{\partial U}{\partial q_1}}{\frac{\partial U}{\partial q_2}}$$

$$q_1$$ is quantity of good 1, and $$q_2$$ is quantity of good 2. U and u are both functions, and it may be a typo that one is not capitalized. I'm not sure, because I'm not sure how to take the partial of that.
 
  • #4
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U and u should be separate functions, I believe, since one of them (U) has two arguments, and the other (u) has only one argument. Since ##U(q_1, q_2) = u(q_1) + q_2##, ##U_1## (which is ##\frac{\partial U}{\partial q_1}##) = ##\frac{\partial}{\partial q_1}(u(q_1) + q_2)##.

The partial derivative of a function of one variable is the same as the derivative of that function.

BTW, the LaTeX notation I used is # # \frac{\partial U}{\partial q_1} # # (but omit the extra spaces).
 
  • #5
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##U(q_1, q_2) = u(q_1) + q_2##, ##U_1## (which is ##\frac{\partial U}{\partial q_1}##) = ##\frac{\partial}{\partial q_1}(u(q_1) + q_2)##.

The partial derivative of a function of one variable is the same as the derivative of that function..
Okay, I think I'm beginning to understand. I need to take ##\frac{\partial}{\partial q_1} (u(q_1) + q_2)##

However, not knowing what ##u(q_1)## is, the solution would be ##\partial u##

And this would make the overall ##MRS = -\frac{\frac{\partial U}{\partial q_1}}{\frac{\partial U}{\partial q_2}} = -\frac{\partial u}{1} = -\partial u##, since ##\frac{\partial U}{\partial q_2} = 1##, correct?


Also, I didn't mention this earlier, but thank you so much for helping me understand this.
 
  • #6
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Okay, I think I'm beginning to understand. I need to take ##\frac{\partial}{\partial q_1} (u(q_1) + q_2)##

However, not knowing what ##u(q_1)## is, the solution would be ##\partial u##
No, ##\partial u## doesn't mean anything, but your reasoning isn't too far off.
$$ \frac{\partial}{\partial q_1} u(q_1) = \frac{d}{d q_1} u(q_1) = ?$$
Dakarai said:
And this would make the overall ##MRS = -\frac{\frac{\partial U}{\partial q_1}}{\frac{\partial U}{\partial q_2}} = -\frac{\partial u}{1} = -\partial u##, since ##\frac{\partial U}{\partial q_2} = 1##, correct?
What you have for ##\frac{\partial U}{\partial q_2}## is correct, but you need to fix the other one.
Dakarai said:
Also, I didn't mention this earlier, but thank you so much for helping me understand this.
You're welcome!
 
  • #7
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No, ##\partial u## doesn't mean anything, but your reasoning isn't too far off.
$$ \frac{\partial}{\partial q_1} u(q_1) = \frac{d}{d q_1} u(q_1) = ?$$
This is where I'm confused. I'm not sure how to solve that. I suppose the simple answer would be ##u'(q_1)## which would make the overall solution ##-u'(q_1)##?
 
  • #8
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This is where I'm confused. I'm not sure how to solve that. I suppose the simple answer would be ##u'(q_1)## which would make the overall solution ##-u'(q_1)##?
Yes, both of those look good.
 
  • #9
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And this is equal to ##-\frac{du}{dq_1}##, correct? That's the answer I was given. And if it is equal, could you explain why it is?
 
  • #10
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And this is equal to ##-\frac{du}{dq_1}##, correct? That's the answer I was given. And if it is equal, could you explain why it is?
Yes, -u'(q1) is the same as ##-\frac{du}{dq_1}##. They are two different notations that represent the same derivative. The prime notation is often called Newton notation (although he used a dot instead of a prime), and the other is usually called Liebniz notation, after the German mathematician who developed a lot of calculus at the same time as Newton.
 

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