<p>For a variable <span class="math-container">$X$</span>, let <span class="math-container">$dX$</span> denote its total differential. Let <span class="math-container">$k$</span> be a constant, and <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> variables. You'll need the following rules:</p>

<p><span class="math-container">$$dk = 0$$</span></p>

<p>(constant rule),</p>

<p><span class="math-container">$$d(X + Y) = dX + dY$$</span></p>

<p>(sum rule),</p>

<p><span class="math-container">$$d(XY) = Y \cdot dX + X \cdot dY$$</span></p>

<p>(product rule) and</p>

<p><span class="math-container">$$d\left(\frac{X}{Y}\right) = \frac{ Y \cdot dX - X \cdot dY }{ Y^2 }$$</span></p>

<p>(quotient rule). These rules follow immediately from the definition of the total differential. (For more on total differentials, see e.&thinsp;g. chapter 8 of Chiang and Wainwright, <em>Fundamental Methods of Mathematical Economics</em>, 4th ed., McGraw-Hill 2005.)</p>

<p>Now let's start with your equation. By the product rule we have:</p>

<p><span class="math-container">$$\begin{align}
dx &amp; = d\left( \mu^{-1} (g + i + e) \right) \\
&amp; = d\left(\mu^{-1}\right) (g + i + e) + \mu^{-1} d(g + i + e)
\end{align}$$</span></p>

<p>What is <span class="math-container">$d(\mu^{-1})$</span>? Because <span class="math-container">$\mu^{-1} = \frac1\mu$</span>, by the quotient rule,</p>

<p><span class="math-container">$$\begin{align}
d\left(\mu^{-1}\right) &amp; = d\left(\frac1\mu\right) \\[4pt]
&amp; = \frac{ \mu \cdot d1 - 1 \cdot d\mu }{ \mu^2 } \\
&amp; = -\frac{ d\mu }{ \mu^2 }
\end{align}$$</span></p>

<p>where we also made use of the fact that by the constant rule, <span class="math-container">$d1 = 0$</span>. Plug this into the above intermediate result and also apply the sum rule:</p>

<p><span class="math-container">$$\begin{align}
dx &amp; = d\left(\mu^{-1}\right) (g + i + e) + \mu^{-1} d(g + i + e) \\
&amp; = -\frac{ d\mu }{ \mu^2 } (g + i + e) + \mu^{-1} \left( dg + di + de \right) \\
&amp; = -\frac{ d\mu }{ \mu } \cdot \mu^{-1} (g + i + e) + \mu^{-1} \left( dg + di + de \right) \\
&amp; = -\dot\mu \cdot x + \mu^{-1} \left( dg + di + de \right)
\end{align}$$</span></p>

<p>where use was made of the definition of <span class="math-container">$x$</span>, and the fact that <span class="math-container">$\frac{ d\mu }\mu = \dot\mu$</span>. </p>

<p>Now we're almost there. Note that since <span class="math-container">$\frac{ dX }X = \dot{X}$</span> for any variable <span class="math-container">$X$</span>, we also have <span class="math-container">$dX = X \cdot \dot X$</span>; applying this to <span class="math-container">$g$</span>, <span class="math-container">$i$</span> and <span class="math-container">$e$</span> and dividing the entire equation by <span class="math-container">$x$</span> then yields</p>

<p><span class="math-container">$$\begin{align}
\dot x = \frac{ dx }x &amp; = -\dot\mu + \mu^{-1} \cdot \frac{ dg + di + de } x \\
&amp; = -\dot\mu + \mu^{-1} \cdot \frac{ dg }x + \mu^{-1} \cdot \frac{ di }x + \mu^{-1} \cdot \frac{ de }x \\
&amp; = -\dot\mu + \frac{ \mu^{-1} g }x \cdot \dot g + \frac{ \mu^{-1} i }x \cdot \dot i + \frac{ \mu^{-1} e }x \cdot \dot e
\end{align}$$</span></p>

<p>which is the desired result.</p>

CC BY-SA 4.0
2019-08-24T08:39:54.587Z
2019-08-24T09:21:55.077Z
2019-08-24T09:21:55.077Z
chsk
<p>Is there anyone who can help me with a total differentiation exercize. I am starting with the following formula for AD:</p>

<p><span class="math-container">$$x=\mu ^{-1}(g+i+e)$$</span></p>

<p>Where <span class="math-container">$\mu$</span> is the Keynesian multiplier.</p>

<p>And have to obtain the following result (which is obtained by total differentiating and diving through by x):</p>

<p><span class="math-container">$$\dot{x}=-\dot\mu+\frac{\mu ^{-1}g}{x}\cdot \dot g+\frac{\mu ^{-1}i}{x}\cdot \dot i+\frac{\mu ^{-1}e}{x}\cdot \dot e$$</span></p>

<p>Thank you very much to anyone who wants to help me with every step.</p>

<macroeconomics><mathematical-economics><economic-growth><linear-algebra><product-differentiation>
Hep with total differentiation of an AD function
Alessandro
# Hep with total differentiation of an AD function

(Asked by: username_0 on 2019-08-24)

Is there anyone who can help me with a total differentiation exercize. I am starting with the following formula for AD:

$$x=\mu ^{-1}(g+i+e)$$

Where $\mu$ is the Keynesian multiplier.

And have to obtain the following result (which is obtained by total differentiating and diving through by x):

$$\dot{x}=-\dot\mu+\frac{\mu ^{-1}g}{x}\cdot \dot g+\frac{\mu ^{-1}i}{x}\cdot \dot i+\frac{\mu ^{-1}e}{x}\cdot \dot e$$

Thank you very much to anyone who wants to help me with every step.

**Tags:** macroeconomics, mathematical-economics, economic-growth, linear-algebra, product-differentiation

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## 1 answers

### Answer by username_1 on 2019-08-24. Score: 2

For a variable $X$, let $dX$ denote its total differential. Let $k$ be a constant, and $X$ and $Y$ variables. You'll need the following rules:

$$dk = 0$$

(constant rule),

$$d(X + Y) = dX + dY$$

(sum rule),

$$d(XY) = Y \cdot dX + X \cdot dY$$

(product rule) and

$$d\left(\frac{X}{Y}\right) = \frac{ Y \cdot dX - X \cdot dY }{ Y^2 }$$

(quotient rule). These rules follow immediately from the definition of the total differential. (For more on total differentials, see e. g. chapter 8 of Chiang and Wainwright, *Fundamental Methods of Mathematical Economics*, 4th ed., McGraw-Hill 2005.)

Now let's start with your equation. By the product rule we have:

$$\begin{align}
dx & = d\left( \mu^{-1} (g + i + e) \right) \\
& = d\left(\mu^{-1}\right) (g + i + e) + \mu^{-1} d(g + i + e)
\end{align}$$

What is $d(\mu^{-1})$? Because $\mu^{-1} = \frac1\mu$, by the quotient rule,

$$\begin{align}
d\left(\mu^{-1}\right) & = d\left(\frac1\mu\right) \\[4pt]
& = \frac{ \mu \cdot d1 - 1 \cdot d\mu }{ \mu^2 } \\
& = -\frac{ d\mu }{ \mu^2 }
\end{align}$$

where we also made use of the fact that by the constant rule, $d1 = 0$. Plug this into the above intermediate result and also apply the sum rule:

$$\begin{align}
dx & = d\left(\mu^{-1}\right) (g + i + e) + \mu^{-1} d(g + i + e) \\
& = -\frac{ d\mu }{ \mu^2 } (g + i + e) + \mu^{-1} \left( dg + di + de \right) \\
& = -\frac{ d\mu }{ \mu } \cdot \mu^{-1} (g + i + e) + \mu^{-1} \left( dg + di + de \right) \\
& = -\dot\mu \cdot x + \mu^{-1} \left( dg + di + de \right)
\end{align}$$

where use was made of the definition of $x$, and the fact that $\frac{ d\mu }\mu = \dot\mu$.

Now we're almost there. Note that since $\frac{ dX }X = \dot{X}$ for any variable $X$, we also have $dX = X \cdot \dot X$; applying this to $g$, $i$ and $e$ and dividing the entire equation by $x$ then yields

$$\begin{align}
\dot x = \frac{ dx }x & = -\dot\mu + \mu^{-1} \cdot \frac{ dg + di + de } x \\
& = -\dot\mu + \mu^{-1} \cdot \frac{ dg }x + \mu^{-1} \cdot \frac{ di }x + \mu^{-1} \cdot \frac{ de }x \\
& = -\dot\mu + \frac{ \mu^{-1} g }x \cdot \dot g + \frac{ \mu^{-1} i }x \cdot \dot i + \frac{ \mu^{-1} e }x \cdot \dot e
\end{align}$$

which is the desired result.


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