Given $f(x+y)=f(x)f(y), f'(0)=11,f(3)=3$, what is $f'(3)$?

Question

The question is this:

Given \begin{align} f(x+y)&=f(x)f(y)\\ f'(0)&=11\\ f(3)&=3 \end{align} What is $f'(3)$?

And my solution:

On differentiating the equation $f(x+y)=f(x)f(y)$ wrt $x$ we get:

$$f'(x+y)(1+y')=f(x)f'(y)(y') + f'(x)f(y)$$

Now on substituting $x=0$ and $y=3$, we get :

$$f'(3)(1+0)=f(0)f'(3)0 + f(3)f'(0)$$

$$f'(3)=3\times11=33$$

Is this a correct way to solve the problem?

Answer 1

Assuming that $f$ has a derivative in $x=3$, then yes. In order to avoid all confusion induced by $y'$ (as show other comments/answers), you should start with simlpier line:

$$f(x+3)=3f(x)$$

and then take the derivative at $x=0$.

Answer 2

you cannot assume that y' at 0 will be zero

Answer 3

Yes, that is a valid way to use the chain rule to get the answer.

Here is another approach. $$ \begin{align} f'(x) &=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{f(x)f(h)-f(x)f(0)}{h}\\ &=f(x)\lim_{h\to0}\frac{f(h)-f(0)}{h}\\ &=f(x)f'(0) \end{align} $$ Thus, $f'(3)=f(3)f'(0)$.