Let `f(x+y)=f(x)dotf(y)` for all `xa n dydot` Suppose `f(5)=2a n df^(prime)(0)=3.` Find `f^(prime)(5)dot`

Explanation

To solve the problem, we start with the functional equation given:

**Step 1:** Given the equation \( f(x+y) = f(x) \cdot f(y) \), we can deduce that this is a form of the exponential function.

**Step 2:** We know that \( f(5) = 2 \) and \( f'(0) = 3 \). We need to find \( f'(5) \).

**Step 3:** We can express \( f'(5) \) using the definition of the derivative:
\[ f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h} \]

**Step 4:** Using the functional equation, we substitute \( x = 5 \) and \( y = h \):
\[ f(5+h) = f(5) \cdot f(h) \]
Thus, we can rewrite the derivative as:
\[ f'(5) = \lim_{h \to 0} \frac{f(5) \cdot f(h) - f(5)}{h} \]

**Step 5:** Factor out \( f(5) \):
\[ f'(5) = f(5) \cdot \lim_{h \to 0} \frac{f(h) - 1}{h} \]

**Step 6:** The limit \( \lim_{h \to 0} \frac{f(h) - 1}{h} \) is the definition of the derivative at 0, which is \( f'(0) \):
\[ f'(5) = f(5) \cdot f'(0) \]

**Step 7:** Substitute the known values \( f(5) = 2 \) and \( f'(0) = 3 \):
\[ f'(5) = 2 \cdot 3 = 6 \]

Thus, the final answer is:
\[ \boxed{6} \]