Question on increasing and decreasing Functons Let $f: \Bbb R\rightarrow \Bbb R$ be a differentiable function such that $f '(0)=1$ and $f(x+y)=f(x)f(y)$ for all $x\in\Bbb R$.
which of the following is true

*

*Both $f$ and $f '$ are decreasing function

*Both $f$ and $f '$ are increasing function

*$f$ is increasing function but $f '$ is decreasing function

*$f$ is decreasing function and $f '$ is decreasing function

My Attempt:
I tried it by taking $x=y$ and then differentiating.
I also tried by putting both $x$ and $y$ equal to $0$.
 A: Differentiate w.r.t $x$ to get $f'(x+y)=f'(x)f(y)$. Put $x=0$ now to get $f'(y)=f(y)$. This implies $f(y)=ce^{y}$ for some constant $c$ and this also implies that $f'=f$. Can you finish?
(You have to go back to the hypothesis to show that $c$ must be $1$).
A: An even more general result holds in relation to your problem:

Assume $f \colon \mathbb{R} \to \mathbb{R}$ is continuous and satisfies the relation $f(x+y)=f(x)f(y)$ for any $x, y \in \mathbb{R}$. Then there exists a real number $a \in \mathbb{R}$ such that $f(x)=\mathrm{e}^{ax}$ for any $x \in \mathbb{R}$.

Proof. $f$ is a morphism between the additive group $(\mathbb{R}, +)$ and the multiplicative monoid $(\mathbb{R}, \cdot)$, whence by virtue of elementary algebra it must be that $\mathrm{Im}f \subseteq \mathbb{R}^*$ (the image is included in the group of units of the target monoid). Furthermore, for arbitrary $x \in \mathbb{R}$ one has $f(x)=f(2\frac{x}{2})=f\left(\frac{x}{2}\right)^2 \geqslant 0$ which combined with the previous observation means that $f(x)>0$ for any $x \in \mathbb{R}$ and hence $\mathrm{Im}f \subseteq (0, \infty)$. Denote by $g={}_{(0, \infty)|}f$ the corestriction of $f$ to the strictly positive domain.
Let us further introduce the map $h=\mathrm{ln} \circ g$. Since the natural logarithm is a continuous morphism from the multiplicative group $\left(\mathbb{R}^*_{+}, \cdot\right)$ to the additive group $(\mathbb{R}, +)$ and $g$ too is a continuous group morphism, their composition $h$ will also be a continuous group morphism from $(\mathbb{R}, +)$ to itself (in other words an endomorphism of the mentioned additive group).
It is however a classical fact that any continuous endomorphism of the group $(\mathbb{R}, +)$ is of the form $x \mapsto tx$ for a certain $t \in \mathbb{R}$. This means in particular that there exists $a \in \mathbb{R}$ such that $h(x)=ax$ for any $x \in \mathbb{R}$. Recalling the definition of $h$, this entails $f(x)=\mathrm{e}^{ax}$ for any $x \in \mathbb{R}$. $\Box$
Returning to the particular case of your problem and adding the hypothesis that $f'(0)=1$, since $f'(x)=a\mathrm{e}^{ax}$ for any $x \in \mathbb{R}$ you can immediately infer that $a=1$ and thus $f$ is the exponential function of base $\mathrm{e}$.
