It looks like I am making a calculus mistake here, and I am really banging my head against the wall.

Here is my work:


I know I should be getting $dY/dT=-(MPC)/(1-MPC)$, where $MPC=dC/dY$ If this is at all helpful here is a picture of page 295 in 7th edition of Macroeconomics By Mankiw that briefly describes the process.


It is confusing to me that he uses $C'$ for both terms on right-hand side when $C$ is composed of two arguments, so I decided to use Leibniz notation to see what is going on.

Thanks for help! Chris

  • $\begingroup$ Imgur link isn't working for me. $\endgroup$
    – 123
    Sep 5 '16 at 1:01
  • $\begingroup$ Thanks for letting me know. I think it is fixed. I guess you have to "BBCode" link for Stackexchange. $\endgroup$ Sep 5 '16 at 1:26
  • $\begingroup$ Is C a function so C=f(Y-T) or is C supposed to be multiplier by Y-T? $\endgroup$
    – VCG
    Sep 5 '16 at 2:30
  • $\begingroup$ Good question: I am convinced it is C=f(Y-T) in the text. Page 290 of Mankiw calls it a consumption function and it also says that C depends on (disposable income). Also Mankiw tends to use this type of notation. Like to model investment as a function of the interest rate he writes I=I(r). $\endgroup$ Sep 5 '16 at 2:42

So here $C$ is a function with the argument $Y-T$, like $C=f(Y-T)$

So differentiate the equation: $dY=C' * (dY-dT)$

The result immediately follows like in the book.

The reason you don't do the partial derivative is that C has only one argument - while Y and T are in there, it only takes in 1 input.

  • $\begingroup$ Ah, so you are saying treat Y-T as one argument (which totally makes sense, because we could just rename the difference as something else and we would have one argument). Cool! So just to clarify we could think of C' being dC/d(Y-T) then, right? Thanks for your help! $\endgroup$ Sep 5 '16 at 3:01
  • $\begingroup$ Also, does this also imply you can distribute the "d" operator in calculus? I have never come across this before. If you know anything that explains this, that would be great. $\endgroup$ Sep 5 '16 at 3:13
  • $\begingroup$ So you definitely do not distribute it, C'=dC/d(Y-T) , but since we are totally differentiating this, we need to chain rule out the inside arguments so I multiply by the derivative of the inside. $\endgroup$
    – VCG
    Sep 5 '16 at 3:39
  • $\begingroup$ I think I understand. It is really weird seeing a function that is of the form f(x-y) instead of f(x,y), and that is really throwing me off. $\endgroup$ Sep 5 '16 at 3:56
  • $\begingroup$ Ya as you said, just think of Y-T as one thing on the inside, and totally differentiate it on the outside. But having multiple things inside 1 argument is something you do come across a decent amount in econ. $\endgroup$
    – VCG
    Sep 5 '16 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.