Let f(x) and g(x) be functions s.t.

f(a+b)=f(a)g(b)+g(a)f(b) for all a,b in R

Given that f(0)=0,f'(0)=1,g(0)=1 and g'(0)=0, show that

f'(x)=g(x) for all x in R

Would appreciate any hints that could be given.

Thanks

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- Sep 27th 2008, 06:14 AMHweengeeStuck bad on this question
Let f(x) and g(x) be functions s.t.

f(a+b)=f(a)g(b)+g(a)f(b) for all a,b in R

Given that f(0)=0,f'(0)=1,g(0)=1 and g'(0)=0, show that

f'(x)=g(x) for all x in R

Would appreciate any hints that could be given.

Thanks - Sep 27th 2008, 07:58 AMShowcase_22
hmm, I tried this ("tried" implies that it fails but you might be able to work some magic on it):

I'm clearly not getting g(x). I think this method is the way to go, except it doesn't use the information f'(0)=1 and g'(0)=0.

I haven't tried induction yet but it works for x=0. This method might take you somewhere. - Sep 27th 2008, 08:20 AMHweengee
i'll try it out, but i've been getting

f'(a+b)=f'(a)g(b)+g'(b)f(a)+g'(a)f(b)+g(a)f'(b)

if b=0,

f'(a+0)=f'(a).1+g(a).1

f'(a)=f'(a)+g(a)

g(a)=0 ?

so maybe the definition of the derivative would be the way to go. - Sep 27th 2008, 08:27 AMMoo
Hello,

I think I got it :)

Assuming the limits exist, blah blah, we have :

and

Hence :

(Tongueout) - Sep 27th 2008, 08:39 AMpankaj
A simple trick is required to prove this.

Do we need values of f(0) and g(0)? - Sep 27th 2008, 08:42 AMHweengee
thanks alot guys. i wonder why differentiating f(a+b) the way i did wouldn't work. could anyone offer some explanation?

thanks. - Sep 27th 2008, 08:57 AMpankaj
O.K. Differentiating implies that you have assumed that the function is differentiable for all x whereas you have been provided values of the derivatives at only 2 points viz. x=a and x=b .That is why you can use only the definition of the derivative.By doing this we are actually checking whether derivatives exist for all x or not if they exist at the given points.

However, had it been given that the function is differentiable I would not have any qualms by proceeding in the way you have but with a slight difference.

Differentiating with respect to treating as a constant

- Sep 27th 2008, 09:04 AMHweengee
yea i get it now. thanks.