A function defined as \( f:R\to R \) satisfies \( f(x+y)=f(x)f(y) \) and \( f(x)\ne0 \) for all \( x,y\in R \), where R is set of real numbers. If \( f(x) \) is differentiable at \( x=0 \) and \( f'(0)=2 \), then \( f'(x) \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2f(x) \)
Putting x = y = 0 gives \( f(0)=f(0)^{2} \). Since \( f(0)\ne0 \), f(0) = 1.
By the definition of the derivative, using \( f(x+h)=f(x)f(h) \),
\( \displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \)
\( \displaystyle =\lim_{h\to0}\frac{f(x)f(h)-f(x)}{h} \)
\( \displaystyle =f(x)\lim_{h\to0}\frac{f(h)-f(0)}{h} \)
\( \displaystyle =f(x)\,f'(0) \)
Since \( f'(0)=2 \),
\( \displaystyle f'(x)=2f(x) \)