# Differentiability of f(x+y)=f(x)*f(y)

**Question**

!!f!! is a function such that !!f(x+y)=f(x)xxf(y)!!, !!f(0)!=0!!. If !!f(x)!! is differentiable at !!x=0!! then for all !!x!! in !!R!!:

- !!f!! is differentiable
- !!f'!! is linear function of !!f!!
- Can't comment

**Solution**

We are talking about differentiablty on !!R!! which is only possible if !!f!! is continuous on !!R!!.

We will first check if !!Lt_(h->0^+) f(x+h)=Lt_(h->0^-) f(x+h)=f(x)!!

Since it is differentiable at !!0!! so it is continuous at !!0!!.

!!=>Lt_(h->0^+) f(0+h)=Lt_(h->0^-) f(0+h)=f(0)!!

More over !!f(0)=f(0)xxf(0)!!. Since !!f(0)!=0!! we get !!f(0)=1!!

Now coming back to !!Lt_(h->0^+) f(x+h)!!

!!=Lt_(h->0^+) f(x)xxf(h)=f(x)Lt_(h->0^+) f(h)=f(x)!!

We can similarly prove !!Lt_(h->0^-) f(x-h)=f(x)!!

Now we look at the differentiablty of !!f(x)!! at any !!x!!

Which is !!Lt_(h->0)(f(x+h)-f(x))/h!!

!!=Lt_(h->0)(f(x)xxf(h)-f(x))/h!!

!!=f(x)Lt_(h->0)(f(h)-1)/h!!

!!=f(x)xxf'(0)!!

So !!f(X) is differentiable for all !!x in R!!

Moreover !!f'!! is a linear function of !!f!!