I know that there is an obvious and tedious way to differentiate it 6 times and then find out the answer but since I have to find the sixth derivative at x=0, I was looking for a short method(Yes, there is surely a method).
I could not find a way to group the trigonometric functions which could make the job easier.
All help is appreciated.
On
Use the following Taylor expansions at $0$. We need the sixth order for $e^x$ and the the trigonometric sum.
$$b(x)=e^x=1+x+{x^2\over 2}+{x^3\over 6}+{x^4\over 24}+{x^5\over 120}+{x^6\over 720}+o(x^6)\\ c(x)=3\sin{x\over\sqrt{3}}+2\cos{\sqrt{3}x}=2+3{x\over\sqrt{3}}-{3x^2}-{x^3\over 6\sqrt{3}}+{3x^4\over 4}+{x^5\over 360\sqrt{3}}-{3x^6\over 40}+o(x^6)\\ $$
We’re interested in the coeffficient of $x^6$ because we know that $a_6=f^{(6)}(0)/6!$ and the coefficient of $x^6$ is
$$a_6=b_1c_5+b_2c_4+b_3c_3+b_4c_2+b_5c_1$$
And surprise surprise one gets $a_6$=0
Hints: $\sin(\theta)=\Im{(e^{i\theta})}$, $\cos(\theta)=\Re{(e^{i\theta})}$, and $\dfrac{d^n}{dx^n}e^{\alpha x} = \alpha^n e^{\alpha x}$.