Use Chain Rule to find `dy/dx," if "y=((3x-1)/(3x+1))^(2)`.

Answer

To find \( \frac{dy}{dx} \) for the function \( y = \left(\frac{3x - 1}{3x + 1}\right)^2 \) using the Chain Rule, we can follow these steps:

### Step 1: Identify the outer and inner functions
Let:
- \( u = \frac{3x - 1}{3x + 1} \) (inner function)
- \( y = u^2 \) (outer function)

### Step 2: Differentiate the outer function
Using the Chain Rule:
\[ \frac{dy}{du} = 2u \]

### Step 3: Differentiate the inner function
To differentiate \( u \), we will use the quotient rule:
\[ u = \frac{a}{b} \quad \text{where } a = 3x - 1 \text{ and } b = 3x + 1 \]
The quotient rule states:
\[ \frac{du}{dx} = \frac{b \cdot \frac{da}{dx} - a \cdot \frac{db}{dx}}{b^2} \]
Calculating \( \frac{da}{dx} \) and \( \frac{db}{dx} \):
- \( \frac{da}{dx} = 3 \)
- \( \frac{db}{dx} = 3 \)

Now substituting into the quotient rule:
\[ \frac{du}{dx} = \frac{(3x + 1)(3) - (3x - 1)(3)}{(3x + 1)^2} \]
Simplifying the numerator:
\[ = \frac{(9x + 3) - (9x - 3)}{(3x + 1)^2} = \frac{6}{(3x + 1)^2} \]

### Step 4: Apply the Chain Rule
Now, using the Chain Rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \frac{du}{dx} \]
Substituting \( u \) and \( \frac{du}{dx} \):
\[ \frac{dy}{dx} = 2\left(\frac{3x - 1}{3x + 1}\right) \cdot \frac{6}{(3x + 1)^2} \]

### Step 5: Simplify the expression
\[ \frac{dy}{dx} = \frac{12(3x - 1)}{(3x + 1)^3} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{12(3x - 1)}{(3x + 1)^3} \]

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