Let `f(x+y)=f(x)+f(y)+x^2y+y^2x` and `lim_(x rarr 0)(f(x))/x=1`. Find `f'(3)`.

Answer

To find \( f'(3) \) given the functional equation \( f(x+y) = f(x) + f(y) + x^2y + y^2x \) and the limit condition \( \lim_{x \to 0} \frac{f(x)}{x} = 1 \), we can follow these steps:

### Step 1: Analyze the Functional Equation

We start with the functional equation:
\[ f(x+y) = f(x) + f(y) + x^2y + y^2x \]
This suggests that \( f(x) \) might be a polynomial function.

### Step 2: Set Up for Derivation

To find the derivative, we can rearrange the functional equation:
\[ f(x+y) - f(x) = f(y) + x^2y + y^2x \]
Now, let’s denote \( y = h \) and analyze the limit as \( h \to 0 \):
\[ f(x+h) - f(x) = f(h) + x^2h + h^2x \]

### Step 3: Divide by \( h \)

Dividing both sides by \( h \):
\[ \frac{f(x+h) - f(x)}{h} = \frac{f(h)}{h} + x^2 + hx \]

### Step 4: Take the Limit as \( h \to 0 \)

Taking the limit as \( h \to 0 \):
\[ f'(x) = \lim_{h \to 0} \left( \frac{f(h)}{h} + x^2 + hx \right) \]
From the limit condition \( \lim_{h \to 0} \frac{f(h)}{h} = 1 \), we substitute:
\[ f'(x) = 1 + x^2 \]

### Step 5: Calculate \( f'(3) \)

Now we need to find \( f'(3) \):
\[ f'(3) = 1 + 3^2 = 1 + 9 = 10 \]

Thus, the final answer is:
\[ \boxed{10} \]