<p>$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\dd}{\, \mathrm{d}}$
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. </p>

<p>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.</p>

CC BY-SA 3.0
2016-02-13T21:12:19.430Z
2016-02-13T21:12:19.430Z
jmbejara
<p>We are given that $C$ is a function of $Y_D$ and $Y_D=Y-Y\tau$. What would be the total differential of $Y=C(Y_D)$?</p>

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

<p>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?</p>

CC BY-SA 3.0
2016-02-13T20:51:24.470Z
Does the function $C$ only have one argument? If so, what is the difference between $C_Y$ and $C_\tau$?
jmbejara
CC BY-SA 3.0
2016-02-13T20:53:50.030Z
$C_Y$ would be partial derivative of consumption C with respect to income Y and $C_\tau$ would be partial derivative with respect to $\tau$
cpage
CC BY-SA 3.0
2016-02-13T21:12:29.317Z
I have replied in my answer.
jmbejara
<macroeconomics><mathematical-economics>
Total Derivative and Composition
cpage
# Total Derivative and Composition

(Asked by: username_0 on 2016-02-13)

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?

**Tags:** macroeconomics, mathematical-economics
> **Comments:**
> **username_1**: Does the function $C$ only have one argument? If so, what is the difference between $C_Y$ and $C_\tau$?  
> **username_0**: $C_Y$ would be partial derivative of consumption C with respect to income Y and $C_\tau$ would be partial derivative with respect to $\tau$  
> **username_1**: I have replied in my answer.  

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## 1 answers

### Answer by username_1 on 2016-02-13. Score: 1 | Accepted answer

$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\dd}{\, \mathrm{d}}$
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 &= \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.


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