# Total Derivative and Composition

We are given that $C$ is a function of $Y_D$ and $Y_D=Y-Y\tau$. What would be the total differential of $Y=C(Y_D)$?

So far I have the following: $$dY=C_{Y_D}(1-\tau)dY+C_{Y_D}(-Y)d\tau$$

However I am not sure how to write this expression in terms of $C_Y$ and $C_{\tau}$. Where do I go from here? Is leaving the expression in terms of $C_{Y_D}$ appropriate?

• Does the function $C$ only have one argument? If so, what is the difference between $C_Y$ and $C_\tau$? – jmbejara Feb 13 '16 at 20:51
• $C_Y$ would be partial derivative of consumption C with respect to income Y and $C_\tau$ would be partial derivative with respect to $\tau$ – cpage Feb 13 '16 at 20:53
• I have replied in my answer. – jmbejara Feb 13 '16 at 21:12

$\newcommand{\pd}{\frac{\partial #1}{\partial #2}} \newcommand{\dd}{\, \mathrm{d}}$ How many arguments does $C$ have? If the only argument to $C$ is $Y_D$, then $C_Y$ and $C_\tau$ might not make sense. I think it's easy to get yourself in trouble here.
Suppose that $C$'s only argument is $Y_D$. Yet, we want to impose the restriction that $Y_D = Y(1-\tau)$. Well, as you wrote \begin{align*} \dd C = C_{Y_D} \dd Y = C_{Y_d} (1-\tau) \dd Y - C_{Y_D} Y \dd \tau \end{align*} is correct. But what is $C_Y$ and $C_\tau$? Well, technically, $$\pd C Y = \pd C \tau = 0.$$ This is because $Y$ and $\tau$ do not directly affect $C$. I guess what you have in mind is something similar to the following. Suppose you define a function $f$ such that $$f(Y, \tau) = C(Y(1-\tau)).$$ Then \begin{align*} f_Y &= \pd f Y = \pd C {Y_D} (1-\tau) \\ f_\tau &= \pd f \tau = -\pd C {Y_D} Y. \end{align*} So, really, $C_Y \neq f_Y$ and $C_\tau \neq f_\tau$, because $C_Y$ and $C_\tau$ are both equal to zero.