Key Concepts and Formulas

• Functional Equation: A functional equation is an equation that relates different values of a function.
• Partial Differentiation: Differentiating a function of multiple variables with respect to one variable, treating other variables as constants.
• Definition of the Derivative: The derivative of a function f(x) at a point is defined as $f'(x) = \mathop {\lim }\limits_{h \to 0} {{f\left( {x + h} \right) - f\left( x \right)} \over h}$.
• L'Hôpital's Rule: If a limit of a quotient of two functions $\mathop {\lim }\limits_{x \to c} {{g\left( x \right)} \over {h\left( x \right)}}$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit is equal to $\mathop {\lim }\limits_{x \to c} {{g'\left( x \right)} \over {h'\left( x \right)}}$, provided the latter limit exists.

Step-by-Step Solution

Step 1: Analyze the given functional equation. We are given the functional equation $f(x+y) = f(x) + f(y) + xy^2 + x^2y$ for all real $x$ and $y$. This equation relates the value of the function at $x+y$ to its values at $x$ and $y$, along with an additional term.

Step 2: Differentiate the functional equation partially with respect to $x$. To find information about the derivative $f'(x)$, we differentiate the given functional equation partially with respect to $x$, treating $y$ as a constant. $\frac{\partial}{\partial x} f(x+y) = \frac{\partial}{\partial x} (f(x) + f(y) + xy^2 + x^2y)$ Using the chain rule on the left side, we get $f'(x+y) \cdot \frac{\partial}{\partial x}(x+y) = f'(x+y) \cdot 1 = f'(x+y)$. On the right side, differentiating term by term: $\frac{\partial}{\partial x} f(x) = f'(x)$ $\frac{\partial}{\partial x} f(y) = 0 \quad (\text{since } y \text{ is treated as a constant})$ $\frac{\partial}{\partial x} (xy^2) = y^2 \quad (\text{since } y^2 \text{ is a constant multiplier})$ $\frac{\partial}{\partial x} (x^2y) = 2xy \quad (\text{since } y \text{ is a constant multiplier})$ Combining these, we get: $f'(x+y) = f'(x) + y^2 + 2xy$

Step 3: Simplify the differentiated equation by substituting specific values for $x$ and $y$. The equation from Step 2 is $f'(x+y) = f'(x) + y^2 + 2xy$. We need to find $f'(0)$ to proceed. Let's substitute $x=0$ into this equation. $f'(0+y) = f'(0) + y^2 + 2(0)y$ $f'(y) = f'(0) + y^2$ This equation now expresses the derivative $f'(y)$ in terms of $f'(0)$ and $y^2$. We can replace $y$ with $x$ to get the general form of the derivative: $f'(x) = f'(0) + x^2 \quad (*).$

Step 4: Determine the value of $f(0)$ from the original functional equation. Let's substitute $x=0$ and $y=0$ into the original functional equation $f(x+y) = f(x) + f(y) + xy^2 + x^2y$. $f(0+0) = f(0) + f(0) + (0)(0)^2 + (0)^2(0)$ $f(0) = 2f(0) + 0 + 0$ $f(0) = 2f(0)$ Subtracting $f(0)$ from both sides, we get: $0 = f(0)$ So, $f(0) = 0$.

Step 5: Use the given limit to find the value of $f'(0)$. We are given the limit $\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$. This limit is in the indeterminate form $\frac{f(0)}{0} = \frac{0}{0}$ (since we found $f(0)=0$). Therefore, we can apply L'Hôpital's Rule. $\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = \mathop {\lim }\limits_{x \to 0} {{{d} \over {dx}} f\left( x \right)} \over {{{d} \over {dx}} x} = \mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over 1}$ We are given that this limit is equal to 1. $\mathop {\lim }\limits_{x \to 0} f'\left( x \right) = 1$ Since $f(x)$ is differentiable, $f'(x)$ is continuous at $x=0$. Thus, $\mathop {\lim }\limits_{x \to 0} f'\left( x \right) = f'(0)$. $f'(0) = 1$

Step 6: Substitute the value of $f'(0)$ back into the expression for $f'(x)$. From Step 3, we have the expression for the derivative: $f'(x) = f'(0) + x^2$. Now, substituting $f'(0) = 1$, we get: $f'(x) = 1 + x^2$

Step 7: Calculate $f'(3)$. We need to find the value of $f'(3)$. Using the expression for $f'(x)$ found in Step 6: $f'(3) = 1 + (3)^2$ $f'(3) = 1 + 9$ $f'(3) = 10$

Common Mistakes & Tips

• Incorrect Differentiation: Ensure partial differentiation is done correctly, treating the other variable as a constant. For example, when differentiating $x^2y$ with respect to $x$, remember to treat $y$ as a constant.
• Confusing $f(x)$ and $f'(x)$: The limit condition $\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$ directly relates to $f'(0)$ via L'Hôpital's rule. Do not confuse this with the original function $f(x)$.
• Algebraic Errors: Simple algebraic mistakes can lead to the wrong final answer. Double-check calculations, especially when substituting values.

Summary

The problem involves a functional equation and a limit condition to find the derivative of the function at a specific point. We began by differentiating the functional equation partially with respect to $x$ and then setting $x=0$ to obtain a general form for $f'(x)$ in terms of $f'(0)$. The given limit, after applying L'Hôpital's Rule, allowed us to determine the value of $f'(0)$. Finally, substituting this value back into the expression for $f'(x)$ gave us the complete form of the derivative, enabling us to calculate $f'(3)$.

The final answer is \boxed{10}.