$$1\\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ \\$$ I've seen a video which was showing a derivative trick in such a way as to make it easier. Consider the pascal's triangle as shown below. On the condition that we have to find $(x^5 e^x )'''$, which is the third derivative, we'll be using these coefficients $1- 4 - 6 -4-1$ and taking derivatives of the first function.
$$x^5 + \color{blue}4\times 5x^4 + \color{blue}6\times 20x^3 + \color{blue}4\times 60x^2+120x$$
Finally, I'll multiply by the derivative of $e^x$ and start from the right.
$$\underbrace{x^5}_{e^x} + 4\times \underbrace{5x^4}_{e^x} + 6\times \underbrace{20x^3}_{e^x} + 4\times \underbrace{60x^2}_{e^x}+\underbrace{120x}_{ e^x}$$
Which yields after factoring
$$e^x(x^5+20x^4+120x^3+240x^2+120x)$$
What if it was $(\cos(x) e^x + \sin(x))'''$? I could have used the same strategy for $\cos(x) e^x$. However, I don't know how to handle $+\sin x $ because it's not product.
Regards
Apply the sum rule repeatedly to obtain
$$ (\mathrm e^x\cos x + \sin x)'''=(\mathrm e^x\cos x)''' + (\sin x)'''\;. $$