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Thread: Derivatives help for economics (PED using calculus = dQ/dP x P/Q)

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    Unhappy Derivatives help for economics (PED using calculus = dQ/dP x P/Q)

    Hi guys, I know this will be a really silly Q but I just cannot understand it.

    For any of you that know some economics, I understand that PED using calculus = dQ/dP x P/Q
    Q = 60 - 3P and so am I right in thinking that the derivative of Q (dQ) is equal to -3?
    We are asked to find the PED when price (P) is equal to 15 so am I also write in thinking that the derivative of P (dP) is 0? According to the constant rule, the derivative of any constant is always 0.
    Therefore, surely dQ/dP is the same as -3/0? and I know you cannot divide by 0?
    the is really bugging me and I know I'm probably making a very silly mistake but I hope someone can help me out.
    Also the book I'm using says that dQ/dP is -3 which is confusing because does that mean they're are just disregarding the fact that the derivative of 15 = 0?

    Many thanks,

    Student

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    Quote Originally Posted by dylanpc View Post
    Hi guys, I know this will be a really silly Q but I just cannot understand it.

    For any of you that know some economics, I understand that PED using calculus = dQ/dP x P/Q
    Q = 60 - 3P and so am I right in thinking that the derivative of Q (dQ) is equal to -3?
    We are asked to find the PED when price (P) is equal to 15 so am I also write in thinking that the derivative of P (dP) is 0? According to the constant rule, the derivative of any constant is always 0.
    Therefore, surely dQ/dP is the same as -3/0? and I know you cannot divide by 0?
    the is really bugging me and I know I'm probably making a very silly mistake but I hope someone can help me out.
    Also the book I'm using says that dQ/dP is -3 which is confusing because does that mean they're are just disregarding the fact that the derivative of 15 = 0?

    Many thanks,

    Student
    What is PED?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    Quote Originally Posted by Subhotosh Khan View Post
    What is PED?

    PED is price elasticity of demand. It's an economics concept but you don't need to know it for the question really I don't think

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    Question hope this make my question more simple

    If this is more simple:

    what is the derivative of the equation Q = 60 - 3P? its -3 right?
    and then what is the derivative of 15? 0?
    so surely you can't work out dQ/dP as this means dividing by 0?

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    Quote Originally Posted by dylanpc View Post
    For any of you that know some economics, I understand that PED using calculus = dQ/dP x P/Q
    Q = 60 - 3P and so am I right in thinking that the derivative of Q (dQ) is equal to -3?
    We are asked to find the PED when price (P) is equal to 15 so am I also write in thinking that the derivative of P (dP) is 0? According to the constant rule, the derivative of any constant is always 0.
    Therefore, surely dQ/dP is the same as -3/0? and I know you cannot divide by 0?
    the is really bugging me and I know I'm probably making a very silly mistake but I hope someone can help me out.
    Also the book I'm using says that dQ/dP is -3 which is confusing because does that mean they're are just disregarding the fact that the derivative of 15 = 0?
    I think you are thoroughly misunderstanding the notation. When we write dQ/dP, it is not a division; it just means the derivative of Q with respect to P. You should think of it as a single symbol. Other books might avoid this notation and call it Q' instead.

    dQ is not the derivative of Q; it is called a "differential", and is not needed here.

    So if Q = 60 - 3P, then we say that dQ/dP = -3, not that dQ = -3. (Note also that we take the derivative of a function, not of an equation, and you must always state with respect to what.)

    Furthermore, you must take a derivative before assigning a specific value to any variable, because once you assign a value, nothing is varying any more, and derivatives are all about how something varies. So if P were a function of something and you took the derivative "when P = 15", you would not take the derivative of the function P = 15, which is 0; you would find the derivative of that function, and then replace P with 15.

    But here P is the independent variable; it is not a function of anything, so you don't take its derivative. You take the derivative of Q with respect to P.

    So, yes, this is a silly question; has it been a long time since you took calculus, or have you not yet really studied it, and are trying to understand the economics with no real knowledge of what derivatives are? We might have suggestions for what you need to study.

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    Quote Originally Posted by dylanpc View Post
    Hi guys, I know this will be a really silly Q but I just cannot understand it.

    For any of you that know some economics, I understand that PED using calculus = dQ/dP x P/Q
    Q = 60 - 3P and so am I right in thinking that the derivative of Q (dQ) is equal to -3?
    We are asked to find the PED when price (P) is equal to 15 so am I also write in thinking that the derivative of P (dP) is 0? According to the constant rule, the derivative of any constant is always 0.
    Therefore, surely dQ/dP is the same as -3/0? and I know you cannot divide by 0?
    the is really bugging me and I know I'm probably making a very silly mistake but I hope someone can help me out.
    Also the book I'm using says that dQ/dP is -3 which is confusing because does that mean they're are just disregarding the fact that the derivative of 15 = 0?

    Many thanks,

    Student
    PED means, I guess, the price elasticity of demand. Are you you too lazy to explain what an acronym means? There are many different types of elasticity.

    It is not true in general that Q = 60 - 3P. Who told you that it was? There is no reason to believe, and quite a few reasons to disbelieve, that demand curves are linear? What evidence do you have for the universal truth that all demand curves are linear with slope equal to - 3. Or is this supposed to be a specific case?

    Yes, in the specific case when [tex]Q = 60 - 3P \implies \dfrac{dQ}{dP} = -\ 3 \implies dQ = -\ 3 dP.[/tex]

    The idea that you can apply calculus to economics in a way that makes sense to people who do not understand calculus is goofy. Of course you cannot divide by zero. But that says nothing about limits, which, according to standard analysis, is what calculus is all about. Of course, most economists are abysmally ignorant of mathematics. How the POINT price elasticity of demand is defined is:

    [tex]Q \ne 0, \text { price elasticity of demand } = \dfrac{dQ}{dP} \div \dfrac{Q}{P}.[/tex]

    It is utter ignorance of modern mathematics that mis-states that as [tex]\dfrac{Q}{dQ} \div \dfrac{P}{dP}.[/tex]

    The average economist has no idea what the mathematical notation means.

    [tex]P = 15 \text { and } \dfrac{dQ}{dP} \div \dfrac{Q}{P} = \dfrac{-\ 3}{\dfrac{60 - 3P}{P}} = \dfrac{-\ 3P}{60 - 3P} = \dfrac{- 3 * 15}{60 - 3 * 15} = -\ 1.[/tex]
    Last edited by JeffM; 12-10-2017 at 12:11 AM.

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    Thanks

    Quote Originally Posted by JeffM View Post
    PED means, I guess, the price elasticity of demand. Are you you too lazy to explain what an acronym means? There are many different types of elasticity.

    It is not true in general that Q = 60 - 3P. Who told you that it was? There is no reason to believe, and quite a few reasons to disbelieve, that demand curves are linear? What evidence do you have for the universal truth that all demand curves are linear with slope equal to - 3. Or is this supposed to be a specific case?

    Yes, in the specific case when [tex]Q = 60 - 3P \implies \dfrac{dQ}{dP} = -\ 3 \implies dQ = -\ 3 dP.[/tex]

    The idea that you can apply calculus to economics in a way that makes sense to people who do not understand calculus is goofy. Of course you cannot divide by zero. But that says nothing about limits, which, according to standard analysis, is what calculus is all about. Of course, most economists are abysmally ignorant of mathematics. How the POINT price elasticity of demand is defined is:

    [tex]Q \ne 0, \text { price elasticity of demand } = \dfrac{dQ}{dP} \div \dfrac{Q}{P}.[/tex]

    It is utter ignorance of modern mathematics that mis-states that as [tex]\dfrac{Q}{dQ} \div \dfrac{P}{dP}.[/tex]

    The average economist has no idea what the mathematical notation means.

    [tex]P = 15 \text { and } \dfrac{dQ}{dP} \div \dfrac{Q}{P} = \dfrac{-\ 3}{\dfrac{60 - 3P}{P}} = \dfrac{-\ 3P}{60 - 3P} = \dfrac{- 3 * 15}{60 - 3 * 15} = -\ 1.[/tex]
    I guess I was too lazy and I just assumed that people would understand the acronym but in future I will explain. And yes I should have specified that this was a specific case where the demand curve is linear. Thanks very much for your time and help!

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    Quote Originally Posted by Dr.Peterson View Post
    I think you are thoroughly misunderstanding the notation. When we write dQ/dP, it is not a division; it just means the derivative of Q with respect to P. You should think of it as a single symbol. Other books might avoid this notation and call it Q' instead.

    dQ is not the derivative of Q; it is called a "differential", and is not needed here.

    So if Q = 60 - 3P, then we say that dQ/dP = -3, not that dQ = -3. (Note also that we take the derivative of a function, not of an equation, and you must always state with respect to what.)

    Furthermore, you must take a derivative before assigning a specific value to any variable, because once you assign a value, nothing is varying any more, and derivatives are all about how something varies. So if P were a function of something and you took the derivative "when P = 15", you would not take the derivative of the function P = 15, which is 0; you would find the derivative of that function, and then replace P with 15.

    But here P is the independent variable; it is not a function of anything, so you don't take its derivative. You take the derivative of Q with respect to P.

    So, yes, this is a silly question; has it been a long time since you took calculus, or have you not yet really studied it, and are trying to understand the economics with no real knowledge of what derivatives are? We might have suggestions for what you need to study.
    Thank you very much for you responses and I understand the problem now. Yes I knew it would be a silly question! and yes it has been 3 years since I've studied any maths at all and then all of a sudden I am doing college/university level maths for economics so the jump has been very large and I would be grateful to hear any suggestions on books you may have to help me with this.

    many thanks,

    student

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    Quote Originally Posted by dylanpc View Post
    Thank you very much for you responses and I understand the problem now. Yes I knew it would be a silly question! and yes it has been 3 years since I've studied any maths at all and then all of a sudden I am doing college/university level maths for economics so the jump has been very large and I would be grateful to hear any suggestions on books you may have to help me with this.

    many thanks,

    student
    It isn't clear whether you are saying that you never studied calculus at all, or that you studied it a little but have forgotten it.

    A course that uses calculus should have a solid knowledge of calculus as a prerequisite; even with that, I would not be surprised if your textbook had an appendix reviewing the essentials of calculus that will be needed in the course. If so, you should start there.

    If I were you, I would be visiting your professor to ask what you will need to know, and how to get that knowledge. The answer will depend on the demands of the course, and on your level of preparation. You should be very open about what you do not understand, in order to get the best possible advice.

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