Finding derivative at a point from functional equation

Question

Let $f(x+y)=f(x)f(y)\forall x,y\in \mathbb{R}$ and $f(5)=2, f'(0)=3$. Then $f'(5)$ is __

I saw this question in a booklet, and based on the answer given, here is what I think they have done.

$$\begin{aligned}f(x+y)&=f(x)f(y)\forall x,y\in \mathbb{R}\\ y=5&\rightarrow f(x+5)=f(x)f(5)\\ &\rightarrow\partial_{x}f(x+5)=\partial_{x}f(x)f(5)\\ &\rightarrow f'(5)=f'(0)f(5)=3\times 2=6\end{aligned}$$

This looks correct, but if we try to write the solutions of the functional equation, which can be $f(x)=0,1,a^{x}(a\gt 0)$, but since $f'(0)\ne 0$, $f(x)=a^{x}$, we should be getting a $\ln()$ term in the derivative, which we do not. So what is wrong here? Thanks.

Answer 1

This seems to be overdetermined. If we have $f(x+y) = f(x)f(y)$, then $f(x) = a^x$. Given that $f(5) = 2$ we can determine that $a^5 = 2 \implies a = 2^{1/5}$, and so $f(x) = 2^{x/5}$. However, now we have $$f'(0) = \ln(2^{1/5})2^{0/5} = \ln(2^{1/5}) \ne 3.$$

Answer 2

Just using the functional relation, for $y \ne 0$, we have that $$ \frac{f(x+y)-f(x)}{y} =\frac{f(x)f(y)-f(x)}{y} = f(x) \cdot \frac{f(y)-1}{y-0} $$

Since $f(0)=1$, and taking the limit as $y \to 0$, we get $$ f'(x) = f(x)\cdot f'(0). $$

In particular, $f'(5) = f(5)\cdot f'(0)$. So, the expected answer would be $f'(5)=6$. As it was pointed out by @DMcMor, another issue is to see if such a function exists...