How do I take high order derivatives for a Taylor expansion? Good afternoon,

I have been assigned to find an 8th order Taylor expansion for an $h(x)=f(x)e^{g(x)}$, where $f(x)$ is a trig function and $g(x)$ is a power of $x$. 

Each derivative expands and after I completed the third derivative I wanted to die. Using the chain and product rule to figure out the 4th may push me over the edge. I am sure there is another way, as my teacher strikes me as far from evil. He also mentioned that the function being even or odd would play a part in some pattern that I may realize... I see that the function is even, but that has not helped me divine any pattern in the subsequent derivatives except that they cycle even/odd.
I wish to become a math wizard, however I am not one currently. For now, I understand basic calculus concepts and operations, so if you can frame your answer with that in mind, it would be immensely appreciated.
I am being vague with the details of the problem here because 1. I don't want to cheat on this assignment (this is for you Prof E., If you are reading this), and 2. I am looking for a general approach to this sort of problem rather than a solution.
Thank you!
 A: Set $u(x)=e^{g(x)}$. Then its derivative satisfies
$$
u'(x)=e^{g(x)}·g'(x)=u(x)·g'(x).
$$
Comparing coefficients you obtain a simple iterative procedure which is essentially a power series product, to compute the Taylor coefficients of $u$ from those of $g$.
At the power $x^{n-1}$ this gives
$$
nu_n=\sum_{k=0}^{n-1}u_k\cdot (n-k)g_{n-k}
$$ 
which allows to successively compute the coefficients $u_n$ from the previously computed, with the initial value $u_0=e^{g_0}$.
Then compute the product of the power series expansions with the usual Cauchy product formula 
$$
h_n=\sum_{k=0}^n f_ku_{n-k}.
$$
Apply factorials to get the derivative values, if necessary.
A: Allow $t(x):=e^{g(x)}$.  Then apply the $8$th derivative of a product:
$$(f\times t)^{(8)}=\sum_{k=0}^8\frac{8!}{k!(8-k)!}f^{(8-k)}\times t^{(k)}$$
Now we apply the $8$th derivative chain rule on $t$, which may be found here.
Substitute that back in above and you get your desired answer.
A: So,
1.) expand e^(-x^2) to get your Taylor Polynomial approximation.
2.) then expand cos(x) as well.
3.) Finally, multiply the two approximations together and combine like terms. 
Now, you have a polynomial that is identical to what you would have gotten if you'd been stubborn enough to expand cos(x)e^(-x^2)!
I'm not sure if it is kosher to answer my own question, but I finished the assignment with some help from the professor and this method was easier to understand for me than the ones offered by the other answers here. Now anyone who has the same question has a few different approaches! 
A: I'll answer for $f(x)=\sin x$ and $g(x)=x^3$.
Multiply 
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}$$ and
$$1+x^3+\frac{x^6}2,$$
keeping only the terms up to degree $8$.

$$x-\frac{x^3}6+x^4+\frac{x^5}{120}-\frac{x^6}6+\frac{2519x^7}{5040}+\frac{x^8}{120}.$$

