Basic calculus/ differentiation

It's linked to the total derivative, I just realised. Not exactly sure how it works, so I can't help, sorry - but you can read on it to see if anything clicks.

Hello,

I am looking over the Mathematical notes and I do not understand how the final equation is obtained, for both note 3 and note 4. When we differentiate Y with respect to G, would we just not get 1 (as G is on it's own?) where does the 1-c come from and why do we divide?

Also, what does the C' mean?

If G is a constant [as the text says] then differentiating a constant gives zero.
dy = C' [dy - dT] expanded gives dy = C' dy - C' dT now rearrange

Rearrange what? I am so confused... Can you please explain to me step by step.

Look, if G is a constant and I differentiate the first equation, do I just not get: dy= c + 0?
Likewise for the second equation, when I differentiate, do I not just get: dy = C(1) - (C*0))?

dY/dT = C' dy/dT - C' dT/dT [dT/dT]

dY/dT - C' dY/dT = - C'

then take out the common factor

Got you, just needed to play around with the algebra. Thanks

C' is shorthand for C differentiated.

and because c is a constant, that is why it is being differentiated right?

C' is shorthand for C differentiated.

C must be something variable - I'm a mathematician not an Economist - is it costs?

(Original post by Muttley79)C must be something variable - I'm a mathematician not an Economist - is it costs?



c is consumption, in this case it is the marginal propensity to consume- so how consumption changes with a change in income

So, to be honest, I don't know what the ' is supposed to indicate. The context of the text (as well as the actual calculations) tend to imply they are treating C as constant. If that's what they are doing, then C' can't sensibly be the derivative (since it would be 0).

So, to be honest, I don't know what the ' is supposed to indicate. The context of the text (as well as the actual calculations) tend to imply they are treating C as constant. If that's what they are doing, then C' can't sensibly be the derivative (since it would be 0).

I freely admit to not being the greatest with partial differentiation, but if C' is supposed to be the derivative, then surely we should have something more like "C'(Y-T) + C dY" (and using dC rather than C' would seem to make more sense as well).

I have to be honest, I'm not 100% trusting the mathematics of the books author.

Edit: I think most mathematically correct would be: $\dfrac{\partial C}{\partial Y}(Y-T)\, dY + C\, dY$, but I'm kind of getting a feeling economists are morbidly afraid of partial derivatives...

Where does it say 'partial; differentiation' - Economics tends to do stuff in a different way and I have no idea what any of the letters represent. I was just trying to help with the rearrangement.