First, use the product rule:
[tex]\dfrac{\mathrm d\left(e^{3x}(\sin(x) + 2\cos(x))\right)}{\mathrm dx} = \dfrac{\mathrm d\left(e^{3x}\right)}{\mathrm dx}(\sin(x)+2\cos(x)) + e^{3x}\dfrac{\mathrm d(\sin(x)+2\cos(x))}{\mathrm dx}[/tex]
For the first term, use the chain rule:
[tex]\dfrac{\mathrm d\left(e^{3x}\right)}{\mathrm dx} = e^{3x}\times\dfrac{\mathrm d(3x)}{\mathrm dx} = e^{3x}\times3 = 3e^{3x}[/tex]
For the second term, use the sum rule:
[tex]\dfrac{\mathrm d(\sin(x)+2\cos(x))}{\mathrm dx} = \cos(x) - 2\sin(x)[/tex]
So we have
[tex]\dfrac{\mathrm d\left(e^{3x}(\sin(x) + 2\cos(x))\right)}{\mathrm dx} = 3e^{3x}(\sin(x)+2\cos(x)) + e^{3x}(\cos(x)-2\sin(x))[/tex]
which can be simplified somewhat as
[tex]\dfrac{\mathrm d\left(e^{3x}(\sin(x) + 2\cos(x))\right)}{\mathrm dx} = e^{3x}\left(3(\sin(x)+2\cos(x))+\cos(x)-2\sin(x)\right) \\\\ \boxed{\dfrac{\mathrm d\left(e^{3x}(\sin(x) + 2\cos(x))\right)}{\mathrm dx} = e^{3x}(\sin(x)+7\cos(x))}[/tex]
