How prove $f(x)$ is a monotonic function if $f(x+y)=f(x)f(y)$ 
Let $f(x)$ be a real valued, differentiable function such that for any $x,y \in \mathbb{R}$,$f(x+y)=f(x)f(y)$. Suppose there exist $a,b$ such that $f(a)\neq 0, f'(b)>0$.
Show that
$f(x)$ is a monotonic function

I tried to use the Cauchy equation result to get: $f(x)=p^x$ for some $p>1$.
I would like to know if there are simpler methods.
 A: Note that $\frac{f(x+h)-f(x)}{h}=f(x)\frac{f(h)-f(0)}{h}$, so $f'(x) = f(x)f'(0)$, and that $f$ is a positive function. 
Putting these two together gives $f'(b) = f(b)f'(0)>0$, so $f'(0)>0$.
Thus for any $x$, $f'(x) \geq 0$ so the function is monotone increasing.
A: Differentiate your equation w.r.t. $x$ and (after that) set $ x=0$. This gives $f'(y) = f'(0) f(y)$ for all $y$ and thus $f(y) = f(0) e^{f'(0) y}$ or all $y$. This should allow you to complete the proof. 
EDIT: Even without ODE, consider
$$
\frac{d}{dx} \frac{f(x)}{e^{f'(0)x}} = \frac{f'(x) e^{f'(0)x} - f(x) f'(0) e^{f'(0)x}}{e^{2f'(0)x}} = 0,
$$
which implies the above formula for $f(y)$. 
A: Watch out:
Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$?
Now assuming $f(x)$ is differentiable and continious we have that only $f(x) = e^{cx}$ statisfy the equation, but $e^{cx}$  strictly increasing and hence strictly monotone.
A: The conditions $f(x+y)=f(x)f(y)$ and $f(a)\neq 0$ imply that $f(x)\neq 0$ for all $x\in\mathbb R$. Since $f$ is continuous, you can assume that $f$ is strictly positive (or consider $-f$ if $f$ is negative instead). Consider then the function $g(x)=\ln(f(x))$, which satisfies
$$g(x+y)=g(x)+g(y),\;\;g'(x)=f'(x)/f(x).$$
Now, $g'(x+y)=\partial_x g(x+y)=g'(x)$, so
$$0<f'(b)/f(b)=g'(b)=g'(b+(x-b))=g'(x)$$
for all $x\in\mathbb R$. In particular, $g$ is strictly increasing, and so is $f=\exp\circ g$.
A: I will give it a try - without the idea that $f(x) = a \exp(x)$...
Given
Let $f(x)$ be a real valued, differentiable function such that for any $x,y \in \mathbb{R} ,f(x+y)=f(x)f(y)$. Suppose there exist $a,b$ such that $f(a) \ne 0$ ,$f'(b) > 0$.
Show that $f(x)$ is a monotonic function

We have
$$\forall\ x \in \mathbb{R} : f(x) \in \mathbb{R},\tag{1}$$
$$\forall\ x,y \in \mathbb{R} : f(x+y) = f(x) f(y),\tag{2}$$
$$\exists\ a \in \mathbb{R} : f(a) \ne 0,\tag{3}$$
$$\exists\ b \in \mathbb{R} : f'(b) > 0.\tag{4}$$

First
$$
f(x) = f\left( \frac{x}{2} + \frac{x}{2} \right) = f^2\left( \frac{x}{2} \right) \ge 0.\tag{5}
$$

Second
$$\Big\{ \exists\ y: f(y) = 0 \Rightarrow \forall\ x: f(x) = f(x-y) f(y) = 0 \Big\}
\tag{6}
$$
$$\Downarrow$$
$$\Big\{ \exists\ y: f(y) \ne 0 \Rightarrow \forall\ x: f(x) \ne 0 \Big\},\tag{7}
$$
whence $f(x) \ne 0$, because $f(a) \ne 0$.
Combination of $(5)$ and $(7)$ yields
$$
f(x) > 0.\tag{8}
$$

Third
$$
f'(x)
= \lim_{y \rightarrow 0} \frac{ f(x + \color{red}{y}) - f(x + \color{blue}{0})}{y}
= f(x) \lim_{y \rightarrow 0} \frac{ f(\color{red}{y}) - f(\color{blue}{0})}{y}
= f(x) f'(0),\tag{9}
$$
whence
$$
f'(0) = \frac{f'(x)}{f(x)}.\tag{10}
$$
As $f'(b) > 0$ and $f(b) > 0$ from $(8)$, we obtain
$$f'(0) > 0.\tag{11}$$

Fourth
As $f(x) > 0 $ from $(8)$ and $f'(0) > 0$ from $(11)$, we obtain
$$
f'(x) > 0,\tag{12}
$$
from $(9)$.

Conclusion:
As $f'(x) > 0$, the function $f(x)$ is monotonic. QED.
