To find the derivative of the function $y = \ln \left( \frac{4x - 7}{x + 7} \right)$, we'll use the chain rule and the quotient rule.

Step 1: Define the inner function

Let $u = \frac{4x - 7}{x + 7}$.

Step 2: Differentiate the natural logarithm function

The derivative of $y = \ln(u)$ with respect to $x$ is: $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$

Step 3: Differentiate the inner function $u = \frac{4x - 7}{x + 7}$

To differentiate $u$, we'll apply the quotient rule, which states: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ Here, $f(x) = 4x - 7$ and $g(x) = x + 7$.

Compute the derivatives: $f'(x) = 4$ $g'(x) = 1$

Apply the quotient rule: $\frac{du}{dx} = \frac{(4)(x + 7) - (4x - 7)(1)}{(x + 7)^2}$

Simplify the numerator: $\frac{du}{dx} = \frac{4x + 28 - 4x + 7}{(x + 7)^2} = \frac{35}{(x + 7)^2}$

Step 4: Substitute back into the derivative of $y$

Now, substitute $u = \frac{4x - 7}{x + 7}$ and $\frac{du}{dx} = \frac{35}{(x + 7)^2}$ into the chain rule formula: $\frac{dy}{dx} = \frac{1}{\frac{4x - 7}{x + 7}} \cdot \frac{35}{(x + 7)^2}$

Step 5: Simplify the expression

$\frac{dy}{dx} = \frac{35(x + 7)}{(x + 7)^2 (4x - 7)} = \frac{35}{(x + 7)(4x - 7)}$

So, the derivative of the function is: $\boxed{\frac{35}{(x + 7)(4x - 7)}}$

Would you like more details on any specific step or have any questions? Here are some related questions to explore further:

1. What is the chain rule in differentiation?
2. How do you apply the quotient rule to a given function?
3. How can we simplify complex fractions in calculus?
4. What is the derivative of the natural logarithm function?
5. How do you interpret the derivative of a logarithmic function graphically?

Tip: When applying the quotient rule, always ensure you correctly identify and differentiate both the numerator and denominator functions before substituting them into the rule.