The problem provides two key conditions about the function $f$:

1. $f(x + y) = f(x) f(y)$ for all real numbers $x$ and $y$.
2. $\lim_{x \to 0} \frac{f(x) - 1}{x} = 5$.

We are asked to use the definition of the derivative to find a relationship between $f'(x)$ and $f(x)$, where the goal is to find a form like:

$f'(x) = \text{some constant} \cdot f(x).$

Step 1: Use the given functional equation

From property (1), the equation $f(x + y) = f(x) f(y)$ suggests that $f$ is an exponential function. This type of functional equation is typically solved by functions of the form $f(x) = e^{kx}$, where $k$ is a constant.

Step 2: Use the limit to find the constant

We can use property (2) to determine the constant $k$. The limit condition is:

$\lim_{x \to 0} \frac{f(x) - 1}{x} = 5.$

Substitute $f(x) = e^{kx}$ into this limit:

$\lim_{x \to 0} \frac{e^{kx} - 1}{x} = 5.$

Using the approximation $e^{kx} \approx 1 + kx$ for small $x$, the limit becomes:

$\lim_{x \to 0} \frac{1 + kx - 1}{x} = k = 5.$

Thus, the constant $k = 5$, and we now have:

$f(x) = e^{5x}.$

Step 3: Derivative of $f(x)$

Now, we differentiate $f(x) = e^{5x}$:

$f'(x) = 5e^{5x} = 5 f(x).$

Final Answer:

The relationship between $f'(x)$ and $f(x)$ is:

$f'(x) = 5 \cdot f(x).$

Would you like further clarification or details?

Here are five related questions to deepen your understanding:

1. What other types of functions satisfy functional equations similar to $f(x + y) = f(x) f(y)$?
2. How does the limit condition help us determine the constant $k$?
3. Can you explain why exponential functions are the typical solution to such functional equations?
4. How would the derivative change if the constant in the exponential function were different from 5?
5. What would happen if the limit condition were different (e.g., the limit was 3 instead of 5)?

Tip: When working with functional equations, look for patterns that suggest common function types, such as exponentials or polynomials.