Thread: Stuck 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

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.

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.

Hello,

I think I got it





Assuming the limits exist, blah blah, we have :



and

Hence :

A simple trick is required to prove this.













Do we need values of f(0) and g(0)?

thanks alot guys. i wonder why differentiating f(a+b) the way i did wouldn't work. could anyone offer some explanation?

thanks.

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

yea i get it now. thanks.