To find the derivative of the given function:

$f(x) = \ln\left( \frac{6x}{x + 9} \right)$

We will apply the chain rule and quotient rule step by step.

Step 1: Use the logarithmic derivative rule

The derivative of $\ln(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$, where $u$ is the inside function. In this case:

$u(x) = \frac{6x}{x + 9}$

So we will first need the derivative of $u(x)$.

Step 2: Differentiate $u(x) = \frac{6x}{x + 9}$ using the quotient rule

The quotient rule states:

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Here, $f(x) = 6x$ and $g(x) = x + 9$.

• The derivative of $f(x) = 6x$ is $f'(x) = 6$.
• The derivative of $g(x) = x + 9$ is $g'(x) = 1$.

Now apply the quotient rule:

$u'(x) = \frac{6(x + 9) - 6x(1)}{(x + 9)^2} = \frac{6x + 54 - 6x}{(x + 9)^2} = \frac{54}{(x + 9)^2}$

Step 3: Apply the chain rule

Now, use the derivative of the natural logarithm function:

$f'(x) = \frac{1}{u(x)} \cdot u'(x)$

Substitute $u(x) = \frac{6x}{x + 9}$ and $u'(x) = \frac{54}{(x + 9)^2}$:

$f'(x) = \frac{1}{\frac{6x}{x + 9}} \cdot \frac{54}{(x + 9)^2} = \frac{x + 9}{6x} \cdot \frac{54}{(x + 9)^2}$

Simplify the expression:

$f'(x) = \frac{54}{6x(x + 9)} = \frac{9}{x(x + 9)}$

Thus, the derivative of the function is:

$f'(x) = \frac{9}{x(x + 9)}$

Would you like further details or have any questions?

Here are some related questions for practice:

1. How would you differentiate a more complex logarithmic function?
2. Can you find the second derivative of $f(x) = \ln\left( \frac{6x}{x + 9} \right)$?
3. How would you apply logarithmic differentiation to a function with powers of $x$?
4. What is the significance of the chain rule when differentiating composite functions?
5. How does the quotient rule differ from the product rule?

Tip: The chain rule is essential when differentiating functions that involve composition, so always identify the "inner" and "outer" functions first.