Let f(x+y)=f(x)*f(y) and f(x)=1+sin(3x)*g(x) where g(x) is a continuous function. Find f '(x) if (f '(0) is not equal to 0) Please note that here $f'$ denotes the derivative of the function $f$.
I tried using the definition of derivative : $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ but that led to a result from which I was not able to conclude anything. Also, I am lacking clarity about partially differentiating multi-variable functions with respect to one variable due to which I could not try that approach.
Any help on the matter is appreciated.
 A: There are actually no functions with these properties.
If $f(x)f(y)=f(x+y),$ then $\log f(x) + \log f(y) = \log f(x+y).$ This is just Cauchy's functional equation. We see that, for all rational $x$ and some fixed $k$, $\log f(x) = kx$, so $f(x)=e^{kx}$ for those $x$. Thus, for all rational $x$,
$$g(x) = \frac{e^{kx}-1}{\sin(3x)}.$$
However, as $x$ gets closer to multiples of $\pi/3$, if $k\neq 0$, $g$ approaches $\pm\infty$, one sign from each side (see, for example, this Desmos plot). Thus, $g$ cannot be continuous if $k\neq 0$. However, if $k=0$, then $f$ is identically $1$, so $f'(0)=0$. 
A: Note that
$$
f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0} \frac{g(x)\sin(3x)}{x}
=3\lim_{x \to 0} \frac{\sin(3x)}{3x}\times \lim_{x \to 0}g(x)=3g(0)
$$
since $g$ is a continuous function and by the well-known limit
$$
\lim_{u \to 0}  \frac{\sin u}{u}=1.
$$
A: Well, you are given that $f$ satisfies the functional equation for exponential function and further that $f(0)=1$ and $f$ is continuous at $0$. It can then be proved with some effort that $f$ is differentiable everywhere and $f(x) =a^x$ for some $a=f(1)>0$. 
Your problem regarding $f'(0)$ is simpler and does not need the functional equation at the beginning (perhaps someone is trying to connect two different unrelated problems, see the other answer which shows that this leads to a contradiction) . Since $f(0)=1$ the value $f'(0)$ is given by limit of $(f(x) - 1)/x$ as $x\to 0$. It follows that $f'(0)=3g(0)$ by the relation given between $f, g$ and the standard limit $(\sin x) /x\to 1$ as $x\to 0$. You don't need the condition that $f'(0)$ should be non-zero. It can be $0$ and there is no problem because of this. 
A: If $f(x+y)=f(x)f(y)$
and $f'$ exists,
then
$f(x) = a^x$
for some $a$.
Therefore
$f(x)
=a^x
=1+\sin(3x)g(x)
$.
Differentiating,
$f'(x)
=a^x\ln a
=\sin(3x)g'(x)+3\cos(3x)g(x)
$.
Setting $x=0$,
$f'(0)
=\ln a
=3g(0)
$
so
$a = e^{3g(0)}$.
Therefore
$f(x)
=e^{3g(0)x}
$.
