Let's treat that as a functional equation
and see what happens.
Suppose
$f(a+b)
=f(a)f'(b)+f(b)f'(a)
$.
My solution,
not quite complete,
is this:
If $f(0) \ne 0$,
$f(x)
=f(0)e^{x/(2f(0))}
$.
If $f(0) = 0$,
$f(x)
=\dfrac{e^{x\sqrt{r}}-e^{-x\sqrt{r}}}{2\sqrt{r}}
$
where
$r = f'''(0)$.
Note:
I show below that,
in this case,
$f'(0) = 1$
and
$f''(0) = 0$,
but haven't been able to show
that
$f'''(0) = -1$.
Once this is shown,
the solution is complete.
Setting $b=0$,
we get
$f(a)
=f(a)f'(0)+f(0)f'(a)
$,
so
$f(0)f'(a)
=f(a)(1-f'(0))
$.
Setting $a=0$ also,
$f(0) = 2f(0)f'(0)$.
If $f(0) \ne 0$,
$f'(0) = \frac12$.
If $f(0) = 0$,
then,
assuming that
$f$ is not constant zero,
$f'(0) = 1$.
If $f(0) \ne 0$,
then
$f'(a)
=cf(a)
$,
where
$c
= \frac{1-f'(0)}{f(0)}
= \frac{1}{2f(0)}
$.
From this,
$(\ln(f(x))' = c$
so
$\ln(f(x))
=cx+d
$,
or
$f(x)
=De^{cx}
$.
For this to satisfy the original equation,
since $f'(x) = cDe^{cx}$,
we want
$De^{c(a+b)}
=De^{ca}(cDe^{cb})+De^{cb}(cDe^{ca})
=2cD^2e^{c(a+b)}
$
so we must have
$2cD = 1$
or
$D
= \frac1{2c}
= \frac1{2(\frac{1}{2f(0)})}
=f(0)
$
(Duh! Obvious from
$f(x) = De^{cx}$).
Therefore,
if $f(0) \ne 0$,
$f(x)
=f(0)e^{x/(2f(0))}
$.
If $f(0) = 0$,
since $f'(0) = 1$,
$f(a+b)
=f(a)f'(b)+f(b)f'(a)
$.
If $b$ is small
and $f'$ and $f''$
are well-behaved,
$f(a+b)
\approx f(a) +bf'(a) + b^2f''(a)/2+b^3f'''(a)/6
$
and
$f(a)f'(b)+f(b)f'(a)
\approx f(a)(f'(0)+bf''(0))+f'(a)(f(0)+bf'(0))
=f(a)+bf''(0)+bf'(a)
$
so $f''(0) = 0$.
Taking an additional term,
$f(a)f'(b)+f(b)f'(a)
\approx f(a)(f'(0)+bf''(0)+b^2f'''(0)/2)+f'(a)(f(0)+bf'(0)+b^2f''(0)/2)
=f(a)(1+b^2f'''(0)/2)+bf'(a)+O(b^3)
$
so
$b^2f''(a)/2
\approx f(a)b^2f'''(0)/2
$
or,
letting $b \to 0$,
$f''(a) = f(a)f'''(0)
$.
Let
$r = f'''(0)$,
so
$f''(a) = rf(a)$
with
$f(0) = 0$
and
$f'(0) = 1$.
The solution to this is
$f(x)
=\dfrac{e^{x\sqrt{r}}-e^{-x\sqrt{r}}}{2\sqrt{r}}
$.
Aha! Looks like $\sin$
if we can show that
$r < 0$
.
However,
I am running out of energy,
so I will stop here.
If someone can finish this,
I will gladly
upvote them.