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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)
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)
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