If f(x + y) = f(x)f(y) for all x, y and f(5) = 2, f'(0) = 3, then f'(5) is (A) 0 (B) 1 (C) 6 (D) 2

Q: Wait, why can I treat $y$ as a constant when differentiating with respect to $x$?

Great question! In the expression $f(x+y)$, $y$ is just a parameter. Since we are taking a partial derivative with respect to $x$, we hold $y$ fixed, effectively treating it as a constant shift. This is a standard and powerful technique in functional equations to isolate variables.

Q: Can I just assume $f(x) = e^{kx}$ and solve it that way?

You absolutely can! Since $f(x+y) = f(x)f(y)$ is the characteristic property of exponential functions, assuming $f(x) = a^x$ or $e^{kx}$ is a brilliant shortcut. However, learning the calculus-based derivation is vital because it works even when the function isn't obviously exponential.

Q: Why did we set $x=0$ specifically?

We set $x=0$ because we were given the value of $f'(0)$. In functional equations, the goal is always to bridge the gap between the general derivative $f'(x+y)$ and the specific information we have at a known point.

Q: Is it possible that $f(x) = 0$ for all $x$?

Mathematically, yes, $f(x)=0$ satisfies the equation. But look at the problem statement: we are given $f(5)=2$. This immediately rules out the trivial zero function. Always check if your given values contradict the trivial solutions!

Q: Do I need to find the explicit form of $f(x)$ to get the answer?

Not at all! The beauty of this problem lies in the fact that we can find the derivative at a specific point without ever needing to know the full algebraic expression for $f(x)$. It is a direct application of the chain rule and the properties of the functional equation.

Question

If $f(x + y) = f(x)f(y) \forall x, y$ and $f(5) = 2, f'(0) = 3$, then $f'(5)$ is

Options:

(A) 0

(B) 1

(C) 6

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we aren't just solving a problem; we are uncovering the hidden symmetry of functions.

When you see an equation like $f(x + y) = f(x)f(y)$, you are looking at the DNA of exponential growth. This specific structure is the hallmark of functions that turn addition into multiplication—a beautiful, deep property that governs everything from compound interest to radioactive decay.

### The Power of Differentiation

We are given the functional equation $f(x + y) = f(x)f(y)$. Our goal is to find $f'(5)$, given $f(5) = 2$ and $f'(0) = 3$.

Many students freeze here, wondering, "How can I find a derivative without knowing the function?" The secret is to stop looking for the function and start looking for the **relationship**.

Let us differentiate both sides of our equation with respect to $x$. Remember, when we differentiate with respect to $x$, we treat $y$ as a constant.

On the left-hand side, we apply the chain rule: the derivative of $f(x+y)$ is simply $f'(x+y)$ multiplied by the derivative of the inner function $(x+y)$ with respect to $x$, which is just $1$. Thus, we get:

$$\frac{d}{dx}[f(x + y)] = f'(x + y)$$

On the right-hand side, since $f(y)$ is treated as a constant, we pull it out of the derivative operator:

$$\frac{d}{dx}[f(x)f(y)] = f'(x)f(y)$$

Equating these two, we arrive at the fundamental identity: $f'(x + y) = f'(x)f(y)$. This is the key that unlocks the entire problem.

### The Strategic Pivot

Now, look at what we have. We have a general derivative $f'(x+y)$ expressed in terms of $f'(x)$ and $f(y)$. We know $f'(0) = 3$.

By setting $x = 0$, we transform our general identity into a specific one:

$$f'(0 + y) = f'(0)f(y)$$

Which simplifies beautifully to:

$$f'(y) = f'(0)f(y)$$

This is a profound result. It tells us that the slope of the function at any point $y$ is directly proportional to the value of the function at that point, with the constant of proportionality being $f'(0)$. This is the very definition of exponential behavior.

### Final Calculation

We are almost there. We need $f'(5)$. Using our derived relation $f'(y) = f'(0)f(y)$, we simply substitute $y = 5$:

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

We were given $f'(0) = 3$ and $f(5) = 2$. Plugging these in, we get:

$$f'(5) = 3 	imes 2 = 6$$

Take a moment to appreciate the elegance of this. We didn't need to solve for $f(x) = e^{3x}$. We didn't need to perform complex integration.

By simply respecting the symmetry of the functional equation and applying the chain rule, the answer revealed itself. The final result is **6**.

Animated Solution for Mathematics - Differentiation: If $f (x + y) = f (x) f (y) \forall x, y$ and $f (5) = 2, f^{'} (0) = 3$ , then $f^{'} (5)$ is

Select Answer:

Visualized Solution

The Functional Equation

Identifying Known Values

Strategy: Partial Differentiation

Differentiating the LHS

Differentiating the RHS

Strategic Substitution

Substituting $x = 0$

Targeting the Goal

Substituting $y = 5$

Evaluating the Final Answer

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Power of Differentiation

The Strategic Pivot

Final Calculation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x+y)=f(x)f(y)∀x,y and f(5)=2,f′(0)=3, then f′(5) is