Basic calculus/ differentiation Watch

Logolept
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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?
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Logolept
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Here is the question
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Chittesh14
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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.
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Logolept
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(Original post by Chittesh14)
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.
I sort of understand it, but I really want the logic to be simplified because I get the equation but only when I work backwards.
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Muttley79
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(Original post by Logolept)
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
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Logolept
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(Original post by Muttley79)
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))?
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Logolept
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Can someone please help me! I am struggling.
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Muttley79
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(Original post by Logolept)
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))?
Let's go back to Y = C{Y - T] + I + G we are told I and G are constant so they differentiate to zero, Yes?

differentiating, dY = etc Y and T are NOT constants

Rearrange dy = C' dy - C' dT
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Muttley79
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dY/dT = C' dy/dT - C' dT/dT [dT/dT = 1]

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

then take out the common factor
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Logolept
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(Original post by Muttley79)
dY/dT = C' dy/dT - C' dT/dT [dT/dT = 1]

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

then take out the common factor
Got you, just needed to play around with the algebra. Thanks
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Muttley79
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(Original post by Logolept)
Got you, just needed to play around with the algebra. Thanks
C' is shorthand for C differentiated.
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Logolept
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(Original post by Muttley79)
C' is shorthand for C differentiated.
and because c is a constant, that is why it is being differentiated right?
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Muttley79
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(Original post by Logolept)
and because c is a constant, that is why it is being differentiated right?
C must be something variable - I'm a mathematician not an Economist - is it costs?
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DFranklin
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(Original post by Muttley79)
C' is shorthand for C differentiated.
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...
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Logolept
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(Original post by Muttley79)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 (C') would be the marginal propensity to consume- so how consumption changes with a change in income
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DFranklin
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(Original post by Logolept)
(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).
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Logolept
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(Original post by DFranklin)
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).
C' is the MPC which is = change in C / change in Y
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pereira325
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(Original post by DFranklin)
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).
' is notation for differentiation. Don't overthink it.
Muttley has given the 'full' mathematical proof
It would not make sense if C is a constant.
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Muttley79
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(Original post by DFranklin)
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.
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pereira325
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(Original post by Muttley79)
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.
It's not partial differentiation.
This is used in 1st year economics in the maths module
(I remember doing it myself)
Partial differentiation comes later and it is quite clear when you see it, as they just do it like normal maths using x,y, etc.
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