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To solve the problem, we need to find the derivative \(\frac{dp}{ds}\) given the function \( p(s) = 3 \ln(4s + 6) + 0.2e^{\cos(s)} \).

### Step 1: Differentiate the first term \( 3 \ln(4s + 6) \).
Using the chain rule:

\[
\frac{d}{ds} \left( 3 \ln(4s + 6) \right) = 3 \cdot \frac{1}{4s + 6} \cdot \frac{d}{ds} (4s + 6)
\]

\[
\frac{d}{ds} (4s + 6) = 4
\]

Thus:

\[
\frac{d}{ds} \left( 3 \ln(4s + 6) \right) = \frac{3 \times 4}{4s + 6} = \frac{12}{4s + 6}
\]

### Step 2: Differentiate the second term \( 0.2e^{\cos(s)} \).
Again, using the chain rule:

\[
\frac{d}{ds} \left( 0.2e^{\cos(s)} \right) = 0.2 \cdot e^{\cos(s)} \cdot \frac{d}{ds} (\cos(s))
\]

\[
\frac{d}{ds} (\cos(s)) = -\sin(s)
\]

So:

\[
\frac{d}{ds} \left( 0.2e^{\cos(s)} \right) = -0.2 \sin(s) e^{\cos(s)}
\]

### Step 3: Combine the derivatives.
Adding the results from Step 1 and Step 2:

\[
\frac{dp}{ds} = \frac{12}{4s + 6} - 0.2 \sin(s) e^{\cos(s)}
\]

This is the derivative of \( p \) with respect to \( s \).

Would you like further details on any of these steps? Here are some related questions you might find useful:

1. How is the chain rule applied in differentiation?
2. What are common derivatives involving logarithmic functions?
3. How does the exponential function's derivative behave?
4. How do trigonometric functions interact with derivatives?
5. How can you simplify expressions after differentiation?

**Tip:** When differentiating composite functions, always apply the chain rule by differentiating the outer function first and then multiplying by the derivative of the inner function.

Determine the derivative dp/ds if p = 3 ln(4s + 6) + 0.2e^cos(s).

Learn how to differentiate p(s) = 3 ln(4s + 6) + 0.2e^cos(s) with respect to s using the chain rule. The problem involves logarithmic, exponential, and trigonometric functions, suitable for advanced high school or college students.

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