Question

Let \( f(x+y) = f(x)f(y) \) and \( f(x) = 1 + \sin(3x)g(x) \), where \( g \) is differentiable. Then \( f'(x) \) is equal to:

• A. \( 3f(x) \)
• B. \( g(0) \)
• C. \( f(x)g(0) \)
• D. \( 3g(x) \)
• E. \( 3f(x)g(0) \)

Hint:

The functional equation \( f(x+y)=f(x)f(y) \) characterizes exponential-type functions \( e^{cx} \). The derivative of such a function is always proportional to the function itself.

Answer 1

Answer: (E)

Concept: For the functional equation \( f(x+y) = f(x)f(y) \), the derivative \( f'(x) \) can be expressed as \( f(x) \cdot f'(0) \). This stems from the definition of the derivative: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f(x) \cdot \lim_{h \to 0} \frac{f(h) - f(0)}{h} \).

Step 1: Determine \( f(0) \).
Substitute \( x=0, y=0 \) into \( f(x+y)=f(x)f(y) \): \[ f(0) = f(0)^2 \implies f(0) = 1 \text{ (as } f(x) \text{ is involving } 1+\sin\dots) \] Also from the definition \( f(x) = 1 + \sin(3x)g(x) \): \[ f(0) = 1 + \sin(0)g(0) = 1 + 0 = 1 \]

Step 2: Calculate \( f'(0) \).
Differentiate \( f(x) = 1 + \sin(3x)g(x) \) using the product rule: \[ f'(x) = \frac{d}{dx}[1] + \frac{d}{dx}[\sin(3x) \cdot g(x)] \] \[ f'(x) = 0 + \cos(3x) \cdot 3 \cdot g(x) + \sin(3x) \cdot g'(x) \] Evaluate at \( x = 0 \): \[ f'(0) = 3\cos(0)g(0) + \sin(0)g'(0) = 3(1)g(0) + 0 = 3g(0) \]

Step 3: Find \( f'(x) \).
Using the functional property \( f'(x) = f(x)f'(0) \): \[ f'(x) = f(x) \cdot 3g(0) = 3f(x)g(0) \]