# Hep with total differentiation of an AD function [closed]

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.

## closed as off-topic by Giskard, jmbejara, Maarten Punt, Bayesian, BB KingSep 18 at 1:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not meet the standards for homework questions as spelled out in the relevant meta posts. For more information, see our policy on homework question and the general FAQ." – Giskard, jmbejara, Maarten Punt, Bayesian, BB King

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.