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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.

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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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