# What rule is used in this example (derivative of a complex function)?

I apologize for the second question on the same day, but I really do need to understand it.

We’ve got a consumption function (it’s from economics but the question is still mathematical) looking like:

C=C(Y−T(Y))C=C(Y-T(Y))

where CC is the consumption, YY is the income and T(Y)T(Y) is the tax as the function of income YY (as in real world the taxes we pay are some proportion of the money we’ve earned);

What we need here is to differentiate this function in respect to Y (e.g. find dCdY\frac {dC}{dY});

When browsing through internet, I’ve actually found a solution for this but I want to figure out how it’s actually done (as it’s a number of problems utilizing the same logic, so if I understand this one, it should be easy to do the rest).

In that solution it’s done the following way

dCdY=dCd(Y−T(Y))∗d(Y−T(Y))dY+dCd(Y−T(Y))∗d(Y−T(Y))dT∗dTdY\frac{dC}{dY} = \frac{dC}{d(Y-T(Y))} * \frac{d(Y-T(Y))}{dY} + \frac{dC}{d(Y-T(Y))} * \frac{d(Y-T(Y))}{dT} * \frac{dT}{dY}

(and then it goes along so that the result looks more elegant)

I wonder which rule is used to do this. It somehow resembles the multivariable version of the chain rule, but the function we’ve got here is quite different from classical examples (like x=y2z3x=y^2z^3 while yy and zz are both some functions). Is it really the chain rule? Or something else?

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It isis the multivariate version of the chain rule.
– Paul
2 days ago

But if we view this function abstractly as C=C(g,h)C=C(g,h) where g and h are both some other functions, what are the gg and hh here? And why are there three steps in the second part after the + sign? I understand the intuition behind that (as we’re “digging” deeper inside the function) but don’t understand the application of the rule in this case
– sempol
2 days ago

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1

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The multivariable chain rule for, for example, a function of the form (t,y(t))↦z(t,y(t)) \mapsto z is:

dzdt=∂z∂t+∂z∂ydydt\dfrac{\mathrm dz}{\mathrm dt} = \dfrac{\partial z}{\partial t} + \dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}

How far you have to go to cover all combinations of derivative of such-and-such with respect to so-and so depends on what the variables implicitly depend on. If you play it safe and go further, and you find you didn’t need to, you should get the right answer anyway.

ProofWiki
Wikipedia

So, the first argument here is YY and we need two steps here (dCd(Y−T(Y))∗d(Y−T(Y))dY\frac{dC}{d(Y-T(Y))} * \frac{d(Y-T(Y))}{dY}) to get to our YY, and the second (complex) argument is Y−T(Y)Y-T(Y) but we can’t get to our YY here in one step, so we need one more step to get from T(Y)T(Y) to the YY inside, right?
– sempol
2 days ago

Yes, and its complexity is that TT depends on YY. And if you use “too many steps” you should get the same answer anyway.
– GFauxPas
2 days ago