If `C(x)= 3ae^(-(x)/(3))` represents the total cost function, find slope of average cost function.

Answer

To solve the problem of finding the slope of the average cost function given the total cost function \( C(x) = 3Ae^{-\frac{x}{3}} \), we will follow these steps:

### Step 1: Define the Average Cost Function
The average cost function \( AC(x) \) is defined as the total cost \( C(x) \) divided by the quantity \( x \):
\[ AC(x) = \frac{C(x)}{x} = \frac{3Ae^{-\frac{x}{3}}}{x} \]

### Step 2: Differentiate the Average Cost Function
To find the slope of the average cost function, we need to differentiate \( AC(x) \) with respect to \( x \):
\[ \frac{d}{dx} AC(x) = \frac{d}{dx} \left( \frac{3Ae^{-\frac{x}{3}}}{x} \right) \]
We will use the quotient rule for differentiation, which states that if \( u = 3Ae^{-\frac{x}{3}} \) and \( v = x \), then:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 3: Compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
- For \( u = 3Ae^{-\frac{x}{3}} \):
\[ \frac{du}{dx} = 3A \cdot \frac{d}{dx} \left( e^{-\frac{x}{3}} \right) = 3A \cdot e^{-\frac{x}{3}} \cdot \left(-\frac{1}{3}\right) = -Ae^{-\frac{x}{3}} \]

- For \( v = x \):
\[ \frac{dv}{dx} = 1 \]

### Step 4: Apply the Quotient Rule
Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:
\[ \frac{d}{dx} AC(x) = \frac{x \cdot \left(-Ae^{-\frac{x}{3}}\right) - 3Ae^{-\frac{x}{3}} \cdot 1}{x^2} \]
This simplifies to:
\[ \frac{d}{dx} AC(x) = \frac{-Ax e^{-\frac{x}{3}} - 3Ae^{-\frac{x}{3}}}{x^2} \]

### Step 5: Factor Out Common Terms
We can factor out \( -Ae^{-\frac{x}{3}} \):
\[ \frac{d}{dx} AC(x) = \frac{-Ae^{-\frac{x}{3}}(x + 3)}{x^2} \]

### Final Result
Thus, the slope of the average cost function is:
\[ \frac{d}{dx} AC(x) = -Ae^{-\frac{x}{3}} \cdot \frac{x + 3}{x^2} \]