Let f(x + y) = f(x)f(y) and f(x) = 1 + sin(3x | vedclass

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(D) Given $f(x+y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$.First,note that $f(0) = f(0+0) = f(0)f(0)$,so $f(0) = 1$ (assuming $f(x) 
eq 0$).Using th

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$3f(x)g(0)$

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(D) Given $f(x+y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$.First,note that $f(0) = f(0+0) = f(0)f(0)$,so $f(0) = 1$ (assuming $f(x) 
eq 0$).Using the definition of the derivative:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$= \lim_{h 	o 0} \frac{f(x)f(h) - f(x)}{h}$$= f(x) \lim_{h 	o 0} \frac{f(h) - 1}{h}$Since $f(h) = 1 + \sin(3h)g(h)$,we have $f(h) - 1 = \sin(3h)g(h)$.Substituting this into the limit:$f'(x) = f(x) \lim_{h 	o 0} \frac{\sin(3h)g(h)}{h}$$= f(x) \lim_{h 	o 0} \left( \frac{\sin(3h)}{3h} 	imes 3 	imes g(h) ight)$$= f(x) 	imes 1 	imes 3 	imes g(0) = 3f(x)g(0)$.Thus,the correct option is $D$.

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

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