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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)

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11 years ago

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11 years ago
```							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)
```
11 years ago
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