We are tasked with differentiating the function:

\[
f(u) = -\frac{a^2}{a + bu} - 2a\ln(a + bu)
\]

with respect to \( u \), where \( a \) and \( b \) are constants.

### Step 1: Differentiate the first term
The first term is \( -\frac{a^2}{a + bu} \). We'll apply the quotient rule to this term, which is:

\[
\frac{d}{du}\left(\frac{f(u)}{g(u)}\right) = \frac{g(u)f'(u) - f(u)g'(u)}{(g(u))^2}
\]

For \( f(u) = a^2 \) (a constant) and \( g(u) = a + bu \), we have:

\[
\frac{d}{du}\left(-\frac{a^2}{a + bu}\right) = -\frac{(a + bu)(0) - a^2(b)}{(a + bu)^2} = \frac{a^2 b}{(a + bu)^2}
\]

### Step 2: Differentiate the second term
The second term is \( -2a\ln(a + bu) \). We'll use the chain rule for logarithmic differentiation:

\[
\frac{d}{du}[\ln(a + bu)] = \frac{1}{a + bu} \cdot b
\]

Thus,

\[
\frac{d}{du}[-2a\ln(a + bu)] = -2a \cdot \frac{b}{a + bu} = -\frac{2ab}{a + bu}
\]

### Step 3: Combine the results
Now, combining the results from both terms, we have:

\[
\frac{d}{du}\left(-\frac{a^2}{a + bu} - 2a\ln(a + bu)\right) = \frac{a^2b}{(a + bu)^2} - \frac{2ab}{a + bu}
\]

This is the derivative of the given expression with respect to \( u \).

Would you like more details or have any questions?

Here are some related questions:
1. How do you apply the quotient rule in more complex functions?
2. What is the chain rule, and how does it work in different contexts?
3. Can we use logarithmic differentiation for other types of functions?
4. How does the derivative of a logarithmic function behave as \( u \) approaches large values?
5. How would the derivative change if the constants \( a \) and \( b \) were dependent on \( u \)?

**Tip:** Always check whether constants or parameters are variable in your problem before applying the rules of differentiation!

Learn how to differentiate the expression -a^2/(a+bu) - 2a ln(a+bu) with respect to u using the quotient and chain rule. This detailed solution breaks down each step, from applying the quotient rule to logarithmic differentiation. Ideal for college-level calculus students.

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