Key Concepts and Formulas

• Functional Equation: The given equation $f(x+y) = f(x)f(y)$ is a characteristic functional equation. Functions satisfying this property are typically of the form $f(x) = a^x$ for some positive constant $a$.
• Differentiation: The process of finding the derivative of a function. The chain rule will be essential here.
• Definition of Derivative: The derivative of a function $f$ at a point $a$, denoted $f'(a)$, can be defined as $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

Step-by-Step Solution

Step 1: Analyze the given functional equation and initial conditions. We are given the functional equation $f(x+y) = f(x)f(y)$ for all $x, y$. We are also given $f(5) = 2$ and $f'(0) = 3$. We need to find $f'(5)$.

Step 2: Differentiate the functional equation with respect to $x$. To find a relationship involving derivatives, we differentiate the given functional equation $f(x+y) = f(x)f(y)$ with respect to $x$, treating $y$ as a constant. Using the chain rule on the left side, $\frac{d}{dx} f(x+y) = f'(x+y) \cdot \frac{d}{dx}(x+y) = f'(x+y) \cdot 1 = f'(x+y)$. On the right side, since $y$ is treated as a constant, $f(y)$ is also a constant. So, $\frac{d}{dx} [f(x)f(y)] = f'(x)f(y)$. Thus, we get: $f'(x+y) = f'(x)f(y)$

Step 3: Substitute specific values to find a relationship between $f'(x)$ and $f(x)$. We know $f'(0) = 3$. Let's substitute $x=0$ into the differentiated equation from Step 2: $f'(0+y) = f'(0)f(y)$ $f'(y) = f'(0)f(y)$ Since $f'(0) = 3$, we have: $f'(y) = 3f(y)$ This relationship holds for all $y$. We can replace $y$ with $x$ to get: $f'(x) = 3f(x)$

Step 4: Calculate $f'(5)$ using the derived relationship. Now we have a direct relationship between the derivative of $f$ at any point $x$ and the value of $f$ at that same point $x$. We need to find $f'(5)$. Using the relationship $f'(x) = 3f(x)$, we can substitute $x=5$: $f'(5) = 3f(5)$ We are given that $f(5) = 2$. Substituting this value: $f'(5) = 3 \times 2$ $f'(5) = 6$

Step 5: Re-examine the problem and the provided solution/options. The provided solution states that the correct answer is A, which corresponds to 0. However, our derivation leads to $f'(5) = 6$. Let's re-evaluate.

Re-evaluation based on the provided correct answer being A (0). If the correct answer is indeed 0, there might be a misunderstanding of the problem or a typo in the question/provided solution. Let's assume for a moment that the problem intends for $f'(5)$ to be 0 and see if any common scenario leads to this.

Consider the form $f(x) = a^x$. Then $f'(x) = a^x \ln a$. Given $f(5) = 2$, we have $a^5 = 2$. Given $f'(0) = 3$, we have $a^0 \ln a = 3$, which means $1 \cdot \ln a = 3$, so $\ln a = 3$. This implies $a = e^3$. Then $f(x) = (e^3)^x = e^{3x}$. Let's check if this satisfies $f(5)=2$: $f(5) = e^{3 \times 5} = e^{15}$. This is not equal to 2. So, $f(x) = a^x$ is not the only function satisfying $f(x+y)=f(x)f(y)$.

Let's go back to the derived relation $f'(x) = f'(0)f(x)$. This relation is robust and correctly derived from the functional equation. $f'(x) = 3f(x)$. Then $f'(5) = 3f(5) = 3 \times 2 = 6$.

There seems to be a contradiction between the derivation and the provided correct answer. Assuming the derivation is correct, the answer should be 6. However, if we are forced to reach the provided correct answer (A) which is 0, we must find a flaw in our reasoning or an alternative interpretation.

Let's consider the possibility that $f(x)$ could be the zero function, i.e., $f(x) = 0$ for all $x$. If $f(x) = 0$, then $f(x+y) = 0$ and $f(x)f(y) = 0 \times 0 = 0$. So $f(x+y) = f(x)f(y)$ is satisfied. However, if $f(x)=0$, then $f(5)=0$, which contradicts $f(5)=2$. Also, $f'(x)=0$ for all $x$, so $f'(0)=0$, which contradicts $f'(0)=3$. Thus, $f(x)=0$ is not the function.

Let's re-examine the differentiation step very carefully. $f(x+y) = f(x)f(y)$ Differentiate with respect to $x$, holding $y$ constant: $f'(x+y) \cdot \frac{d}{dx}(x+y) = f'(x)f(y) + f(x)\frac{d}{dx}(f(y))$ $f'(x+y) \cdot 1 = f'(x)f(y) + f(x) \cdot 0$ (since $f(y)$ is constant with respect to $x$) $f'(x+y) = f'(x)f(y)$

Now, let $x=0$: $f'(y) = f'(0)f(y)$ Given $f'(0)=3$, we have: $f'(y) = 3f(y)$ This implies $f'(x) = 3f(x)$ for all $x$.

Now we need $f'(5)$. Substitute $x=5$: $f'(5) = 3f(5)$ Given $f(5)=2$: $f'(5) = 3 \times 2 = 6$.

It appears there is a discrepancy with the provided correct answer. Based on standard calculus and functional equation properties, the result is 6. If the intended answer is 0, there might be a subtle point missed or an error in the question/options.

However, let's consider if there's a scenario where $f'(5)$ can be 0. If $f(x) = c$ (a constant), then $c = c^2$, so $c=0$ or $c=1$. If $f(x)=0$, $f(5)=0$ (contradicts $f(5)=2$). If $f(x)=1$, $f(5)=1$ (contradicts $f(5)=2$).

Let's assume the question implies that $f(x)$ is a function such that $f(x+y)=f(x)f(y)$. This implies $f(x)=a^x$ or $f(x)=0$. If $f(x)=a^x$: $f(5) = a^5 = 2$. $f'(x) = a^x \ln a$. $f'(0) = a^0 \ln a = \ln a = 3$. So $a = e^3$. Then $f(x) = (e^3)^x = e^{3x}$. $f(5) = e^{15}$, which contradicts $f(5)=2$.

This indicates that the function is not necessarily of the form $a^x$. The derivation $f'(x) = f'(0)f(x)$ is general and does not assume the form $a^x$.

Given the strict instruction to arrive at the provided correct answer, and the strong contradiction derived, it's impossible to logically reach the answer 0 from the given information and standard mathematical principles. The provided solution in the prompt, which also arrives at 6, further reinforces this.

Let's assume there's a mistake in the provided "Correct Answer" and proceed with the derived answer. My derivation consistently leads to $f'(5) = 6$.

Revisiting the prompt's provided solution: The provided solution in the prompt is: $f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$ Differeniate with respect to $x,$ treating $y$ as constant $f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$ Putting $x=0$ and $y=x$, we get $f'\left( x \right) = f'\left( 0 \right)f\left( x \right);$ $\Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.$ This matches my derivation exactly. The prompt states "Correct Answer: A" which is 0. This is a direct contradiction.

Given the rules, I must assume the provided "Correct Answer" (A, which is 0) is correct and work backwards or find a flaw in my reasoning. However, the reasoning is standard.

Let's consider a scenario where the function might behave differently. If $f(x) = e^{kx}$ for some constant $k$. $f(x+y) = e^{k(x+y)} = e^{kx}e^{ky} = f(x)f(y)$. This form works. $f(5) = e^{5k} = 2$. $f'(x) = ke^{kx}$. $f'(0) = ke^0 = k = 3$. So, $k=3$. Then $f(x) = e^{3x}$. $f(5) = e^{15}$. This contradicts $f(5)=2$.

The only way $f'(5)$ could be 0 is if $f'(0)=0$ or $f(5)=0$. Neither is true.

There appears to be an error in the problem statement or the provided correct answer. My derivation consistently yields 6. However, I am compelled to reach the stated correct answer if possible.

Let's assume there's a typo in the question and $f'(0)$ was meant to be something else or $f(5)$ was meant to be something else.

If $f'(5) = 0$, and $f'(x) = f'(0)f(x)$, then either $f'(0)=0$ or $f(5)=0$. If $f'(0)=0$, then $f'(x)=0$ for all $x$, which means $f(x)=c$. $f(5)=c=2$. $f'(0)=0$ contradicts $f'(0)=3$. If $f(5)=0$, then $f'(5) = f'(0) \cdot 0 = 0$. But $f(5)=2$, not 0.

Given the constraints, I cannot logically derive 0 as the answer. The provided solution in the prompt also derives 6. I will proceed with the derived answer, acknowledging the discrepancy. However, if forced to choose option A (0), I would have to hypothesize an error in the problem.

Let's assume, hypothetically, that the problem meant to ask for something else or had different initial conditions that would lead to 0. Without such information, the answer is 6.

Given the strict instruction to match the "Correct Answer", and the provided answer is A (0), but my derivation (and the prompt's own provided solution) yields 6, there is an unresolvable conflict. I must prioritize the derivation. However, if I strictly adhere to reaching the "Correct Answer", I cannot provide a valid step-by-step derivation.

Let me assume there is a very subtle aspect I am missing. The functional equation $f(x+y) = f(x)f(y)$ implies $f(x) = a^x$ for $f(x) \neq 0$. If $f(x)=0$ for some $x_0$, then $f(x_0+y) = f(x_0)f(y) = 0 \cdot f(y) = 0$. This implies $f(z)=0$ for all $z > x_0$. Similarly, $f(y+x_0) = f(y)f(x_0) = 0$. If $f(x)=a^x$, then $f'(x) = a^x \ln a$. $f(5) = a^5 = 2$. $f'(0) = a^0 \ln a = \ln a = 3$. This leads to $a = e^3$. $f(x) = e^{3x}$. $f(5) = e^{15} \neq 2$.

This means the assumption $f(x)=a^x$ is not directly applicable if we use both conditions simultaneously to determine $a$.

The derivation $f'(x) = f'(0)f(x)$ is fundamental. $f'(5) = f'(0)f(5) = 3 \times 2 = 6$.

It is impossible to reach the answer 0 with the given information and standard mathematical methods. The problem statement or the provided correct answer is likely erroneous.

However, I am instructed to produce a solution that leads to the correct answer. If the correct answer is 0, then there must be a scenario where this happens. This requires $f'(5)=0$. From $f'(x) = f'(0)f(x)$, for $f'(5)=0$, we need $f'(0)=0$ or $f(5)=0$. We are given $f'(0)=3 \neq 0$ and $f(5)=2 \neq 0$. Therefore, $f'(5)$ cannot be 0 under these conditions.

Given the constraint that I must arrive at the provided correct answer (A, which is 0), and my derivation consistently leads to 6, I cannot fulfill the request without introducing incorrect mathematics or baseless assumptions. The problem as stated, with the given correct answer, is contradictory.

If I were forced to select option A, I would state that there is an inconsistency.

If, purely hypothetically, we assume the question was designed such that $f'(5)=0$: This would require either $f'(0)=0$ or $f(5)=0$. Since $f'(0)=3$ and $f(5)=2$, this is not possible.

Let me consider if the question implies a specific domain or property that I have overlooked. "$\forall x, y$" suggests it holds for all real numbers.

Final attempt to interpret to reach 0: Could there be a case where $f(x)$ is not differentiable everywhere, but $f'(0)$ exists? The problem states $f'(0)=3$, implying differentiability at 0.

Let's revisit the prompt's solution which yielded 6. This reinforces that 6 is the mathematically derived answer.

Since I must provide a solution that reaches the given correct answer, and it's impossible, I will highlight the discrepancy. However, the instructions forbid meta-commentary.

Given the strict rules, and the impossibility of reaching answer A (0) from the provided information, I cannot proceed to create a valid step-by-step solution that matches the "Correct Answer". The derived answer is 6.

Let's assume there's a typo in the question and $f'(0)=0$. Then $f'(x) = 0 \cdot f(x) = 0$. So $f'(5)=0$. This would lead to option A. However, the question explicitly states $f'(0)=3$.

Let's assume there's a typo and $f(5)=0$. Then $f'(5) = f'(0) \cdot f(5) = 3 \cdot 0 = 0$. This would lead to option A. However, the question explicitly states $f(5)=2$.

Since I cannot logically derive the provided answer, I will present the derivation that leads to 6 and then state the discrepancy. However, the rules prohibit this.

I am in a paradox. I must arrive at the correct answer (A, 0), but the logical derivation leads to 6.

Let's assume the question setter made an error and meant to imply conditions that lead to 0. Without knowing those intended conditions, I cannot construct a valid derivation.

I will proceed with the derivation that yields 6, as it is mathematically sound based on the problem statement. If the correct answer is indeed 0, then the problem statement is flawed.

Step-by-Step Solution

Step 1: Identify the given functional equation and initial conditions. We are given the functional equation $f(x+y) = f(x)f(y)$ for all real numbers $x$ and $y$. We are also given $f(5) = 2$ and $f'(0) = 3$. We need to find the value of $f'(5)$.

Step 2: Differentiate the functional equation with respect to $x$. To find a relationship involving the derivative, we differentiate $f(x+y) = f(x)f(y)$ with respect to $x$, treating $y$ as a constant. Applying the chain rule to the left side: $\frac{d}{dx} f(x+y) = f'(x+y) \cdot \frac{d}{dx}(x+y) = f'(x+y) \cdot 1 = f'(x+y)$. On the right side, since $y$ is constant, $f(y)$ is also constant. Thus, $\frac{d}{dx} [f(x)f(y)] = f'(x)f(y)$. Equating both sides, we get: $f'(x+y) = f'(x)f(y)$

Step 3: Substitute $x=0$ to find a general relation for $f'(x)$. Let $x=0$ in the differentiated equation from Step 2: $f'(0+y) = f'(0)f(y)$ $f'(y) = f'(0)f(y)$ We are given that $f'(0) = 3$. Substituting this value: $f'(y) = 3f(y)$ This relation holds for all real numbers $y$. We can replace $y$ with $x$ to write: $f'(x) = 3f(x)$

Step 4: Calculate $f'(5)$ using the derived relation and given information. Now we have a direct relationship between the derivative of the function at any point $x$ and the value of the function at that point. To find $f'(5)$, we substitute $x=5$ into the relation $f'(x) = 3f(x)$: $f'(5) = 3f(5)$ We are given that $f(5) = 2$. Substituting this value: $f'(5) = 3 \times 2$ $f'(5) = 6$

Common Mistakes & Tips

• Incorrect Differentiation: Ensure the chain rule is applied correctly when differentiating $f(x+y)$ with respect to $x$, and remember that $f(y)$ is treated as a constant in this differentiation.
• Assuming $f(x) = a^x$ prematurely: While the functional equation strongly suggests $f(x)=a^x$, it's best to derive general relations first, as shown in the steps, before assuming a specific form, especially when multiple initial conditions are provided that might lead to contradictions if the form is assumed too early.
• Ignoring Given Conditions: Always use all the given conditions ($f(5)=2$ and $f'(0)=3$) to solve the problem.

Summary The problem involves a functional equation $f(x+y) = f(x)f(y)$, which is characteristic of exponential functions. By differentiating this equation with respect to $x$ and using the given initial conditions $f(5)=2$ and $f'(0)=3$, we derived a general relationship $f'(x) = 3f(x)$. Applying this relation at $x=5$, we found $f'(5) = 3f(5) = 3 \times 2 = 6$.

Note: There is a discrepancy between the derived answer (6) and the provided correct answer (A, which is 0). Based on standard mathematical principles, the derivation leading to 6 is correct. If the intended answer is 0, the problem statement likely contains an error or omission.

The final answer is \boxed{6}.