Understanding the Functional Equation and Derivatives

The given problem involves a functional equation $f(x + y) = f(x) f(y)$ and asks for the value of the derivative of $f(x)$ at a specific point, $x=5$. Functional equations describe properties of functions. The given equation $f(x + y) = f(x) f(y)$ is a well-known property characteristic of exponential functions.

Deriving the Derivative of f(x)

To find $f'(x)$, we use the definition of the derivative:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

We can use the given functional equation $f(x + y) = f(x) f(y)$ by substituting $y = h$. This gives $f(x + h) = f(x) f(h)$.

Substitute this into the derivative definition:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x) f(h) - f(x)}{h}$

Factor out $f(x)$ from the numerator:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x) [f(h) - 1]}{h}$

Since $f(x)$ does not depend on the variable $h$ in the limit, we can pull it outside the limit:

$\displaystyle f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$

Now, let's evaluate the functional equation at $x=0$ and $y=0$: $f(0+0) = f(0)f(0)$, which simplifies to $f(0) = [f(0)]^2$. This equation has two possible solutions for $f(0)$: $f(0) = 0$ or $f(0) = 1$.

• If $f(0)=0$, then from $f(x) = f(x+0) = f(x)f(0) = f(x) \cdot 0 = 0$, we get $f(x)=0$ for all $x$. In this case, $f'(x) = 0$ for all $x$. Then $f'(5)=0$, $f(5)=0$, and $f'(0)=0$. The relationship $f'(5) = f(5) f'(0)$ becomes $0 = 0 \cdot 0$, which is true.
• If $f(0)=1$, consider the expression $\lim_{h \to 0} \frac{f(h) - 1}{h}$. This limit resembles the definition of the derivative of $f(x)$ evaluated at $x=0$: $\displaystyle f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$. Since we are considering the case $f(0)=1$, the limit becomes $\lim_{h \to 0} \frac{f(h) - 1}{h}$, which is exactly $f'(0)$.

In both cases (whether $f(0)=0$ or $f(0)=1$), the limit $\lim_{h \to 0} \frac{f(h) - 1}{h}$ is equal to $f'(0)$.

Therefore, the derivative of $f(x)$ can be expressed as:

$\displaystyle f'(x) = f(x) f'(0)$

Evaluating f'(5)

To find $f'(5)$, we substitute $x=5$ into the derived formula for $f'(x)$:

$\displaystyle f'(5) = f(5) f'(0)$

Comparing with Options

We compare our result with the given options:

Our derived result $f'(5) = f(5) f'(0)$ matches Option 1.

Summary of Steps

1. Start with the definition of the derivative for $f'(x)$.
2. Use the given functional equation $f(x+y)=f(x)f(y)$ to replace $f(x+h)$.
3. Factor out $f(x)$ from the limit expression.
4. Identify the remaining limit as $f'(0)$, considering the two possibilities for $f(0)$ derived from $f(0)=[f(0)]^2$.
5. Substitute $x=5$ into the formula for $f'(x)$.
6. Compare the result with the provided options.

Revision Table: Functional Equation Derivatives

Additional Information: Properties of $f(x+y)=f(x)f(y)$

The functional equation $f(x+y)=f(x)f(y)$ has solutions of the form $f(x) = a^x$ for some constant $a > 0$, or $f(x) = 0$.

• If $f(x) = a^x$, then $f(x+y) = a^{x+y} = a^x a^y = f(x)f(y)$. This confirms $a^x$ is a solution.
• The derivative of $f(x) = a^x$ is $f'(x) = a^x \ln(a)$.
• For $f(x) = a^x$, we have $f'(0) = a^0 \ln(a) = 1 \cdot \ln(a) = \ln(a)$.
• So, $f'(x) = a^x \ln(a) = a^x \cdot f'(0) = f(x) f'(0)$. This matches our derived general formula for the derivative.
• Specifically, $f'(5) = a^5 \ln(a)$. Using the derived formula $f'(5) = f(5) f'(0)$, we get $f'(5) = a^5 \cdot \ln(a)$, which is consistent.
• The other possible solution is $f(x)=0$ for all $x$. As discussed, this also satisfies the functional equation and the relationship $f'(x)=f(x)f'(0)$.

The result $f'(5) = f(5) f'(0)$ holds true for all differentiable functions satisfying the given functional equation.