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?
$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\dd}{\, \mathrm{d}}$
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 &amp;= \pd f Y = \pd C {Y_D} (1-\tau) \\
f_\tau &amp;= \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.