We are given that $f : (0, +\infty) \to \mathbb{R}$ is a function satisfying:

1. $f'(1) = A$ (where $A$ is a constant),
2. For any $x_1, x_2 > 0$, the functional equation $f(x_1 x_2) = x_2 f(x_1) + x_1 f(x_2)$ holds.

We are tasked with finding $f(1)$ and proving that $f'(x_0) = A + \frac{f(x_0)}{x_0}$ for any $x_0 > 0$.

Step 1: Find $f(1)$

Let's set $x_1 = x_2 = 1$ in the functional equation $f(x_1 x_2) = x_2 f(x_1) + x_1 f(x_2)$. This gives: $f(1 \cdot 1) = 1 \cdot f(1) + 1 \cdot f(1),$ which simplifies to: $f(1) = 2f(1).$ This equation implies that: $f(1) = 0.$

Step 2: Prove $f'(x_0) = A + \frac{f(x_0)}{x_0}$

We now aim to prove that $f'(x_0) = A + \frac{f(x_0)}{x_0}$ for any $x_0 > 0$.

Substitution step:

Let $x_1 = x_0$ and $x_2 = y$ in the functional equation $f(x_1 x_2) = x_2 f(x_1) + x_1 f(x_2)$. This gives: $f(x_0 y) = y f(x_0) + x_0 f(y).$ Now, differentiate both sides with respect to $y$. Using the product rule on the right-hand side, we get: $\frac{d}{dy} f(x_0 y) = f(x_0) + x_0 f'(y).$ On the left-hand side, using the chain rule, we have: $f'(x_0 y) \cdot x_0 = f(x_0) + x_0 f'(y).$ Now, set $y = 1$ in this equation: $f'(x_0) \cdot x_0 = f(x_0) + x_0 f'(1).$ Since $f'(1) = A$, we substitute this into the equation: $f'(x_0) \cdot x_0 = f(x_0) + x_0 A.$ Dividing through by $x_0$, we obtain: $f'(x_0) = A + \frac{f(x_0)}{x_0}.$ This completes the proof.

Final Answer:

• $f(1) = 0$,
• $f'(x_0) = A + \frac{f(x_0)}{x_0}$ for any $x_0 > 0$.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. What properties can be deduced about the function $f(x)$ from the functional equation?
2. How does the condition $f'(1) = A$ impact the behavior of $f(x)$?
3. What other techniques can be used to solve functional equations involving derivatives?
4. How does the result $f(1) = 0$ affect the form of the function $f(x)$?
5. Can we determine $f(x)$ explicitly for any special cases?

Tip: When working with functional equations, try substituting simple values or introducing new variables to simplify the given equation.