Q.A function $f : \mathbb{R} \to \mathbb{R}$ satisfies the equation $f(x + y) = f(x)\,f(y)$ for all $x, y \in \mathbb{R}$, $f(x) \ne 0$. Suppose that the function is differentiable at $x = 0$ and $f'(0) = 2$. Prove that $f'(x) = 2f(x)$.

Concept: Functional Equation Derivative — Differentiate the given functional equation with respect to $y$, then use the known derivative at $0$.

Step 1: Fix $x \in \mathbb{R}$. Differentiate $f(x+y) = f(x)f(y)$ with respect to $y$:

Step 2: Set $y = 0$: …

The key idea is that the functional equation $f(x+y)=f(x)f(y)$ is the Cauchy exponential equation. When $f$ is nonzero and differentiable at $0$, it forces $f$ to be of the form $f(x)=e^{kx}$ for some constant $k$. But we don't need to find $f$ explicitly — we can directly compute $f'(x)$ using the definition of the derivative and the functional equation.

Let's work through this step by step.

1. Set up the derivative definition. For any $x \in \mathbb{R}$,

2. Use the functional equation to rewrite $f(x+h)$. Since $f(x+h) = f(x)f(h)$ for all $x, h$, we have

3. Identify the limit as $f'(0)$. We know $f(0) = 1$ because setting $y=0$ in the functional equation gives $f(x) = f(x)f(0)$, and since $f(x) \ne 0$, we get $f(0)=1$. Therefore, $\lim_{h \to 0} \frac{f(h) - 1}{h} = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = f'(0).$ …

Method: Differentiating a Functional Equation at a Known Point

Use this method whenever a function is defined by a relation that must hold for all real $x,y$ — such as $f(x+y)=f(x)f(y)$ — and you are given the derivative at one specific point (often $x=0$) and asked to find a formula for $f'(x)$ everywhere.

Steps

Step 1: Write the derivative of $f$ at a general point using the limit definition

This is the starting point for any functional-equation problem that asks you to prove a formula for $f'(x)$, because it lets you bring the given functional equation directly into the derivative.

Step 2: Substitute the functional equation into $f(x+h)$

Since the relation holds for all inputs, it holds in particular for $f(x+h)$, letting you rewrite it in terms of $f(x)$ and $f(h)$ (here, $f(x+h)=f(x)f(h)$), and factor $f(x)$ out of the limit since it does not depend on $h$.

Step 3: Recognise the remaining limit as the given derivative at the special point …

Common Mistakes

Mistake 1: Forgetting to establish $f(0)=1$ before using it

Why it's wrong: the step that identifies the leftover limit as $f'(0)$ implicitly assumes $f(0)=1$ (so that $\frac{f(h)-1}{h}$ matches $\frac{f(0+h)-f(0)}{h}$); skipping this makes the identification of the limit as $f'(0)$ invalid. Correct approach: always derive $f(0)$ from the functional equation itself (by substituting $y=0$) before using it in the limit manipulation.

Mistake 2: Guessing an explicit formula for $f(x)$ and differentiating that instead …