let f(x+y)=f(x)f(y)and f(x)=1+(sin2x)g(x) where g(x) is continuous than find f'(x)

f(x+y) - f(x) = f(x) [ f(y) - 1]

f(x+y) - f(x)/y =  {f(x) [ f(y) - 1]}/y

lim y tends to 0 f(x+y) - f(x)/y =   f(x) lim y tends to 0 [ f(y) - 1]}/y

f ' (x) = f(x) f ' (0)              [ because from given f(0) = 1] ................................................  (1)

now , f(x) = 1 + sin2x  g(x)

so f ' (x) = 2 cos2x g(x) + sin2x g ' (x)

so  f ' (0) = 2 g(x)

substituting value of f ' (0) in (1) we get

f ' (x)= 2 f(x) g(x)