Question

Can someone differentiate
e^(3x).(sinx +2cosx) !!!

Answer 1

First, use the product rule:

`(\mathrm d\left(e^(3x)(\sin(x) + 2\cos(x))\right))/(\mathrm dx) = (\mathrm d\left(e^(3x)\right))/(\mathrm dx)(\sin(x)+2\cos(x)) + e^(3x)(\mathrm d(\sin(x)+2\cos(x)))/(\mathrm dx)`

For the first term, use the chain rule:

`(\mathrm d\left(e^(3x)\right))/(\mathrm dx) = e^(3x)*(\mathrm d(3x))/(\mathrm dx) = e^(3x)*3 = 3e^(3x)`

For the second term, use the sum rule:

`(\mathrm d(\sin(x)+2\cos(x)))/(\mathrm dx) = \cos(x) - 2\sin(x)`

So we have

`(\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))`

which can be simplified somewhat as

`(\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{(\mathrm d\left(e^(3x)(\sin(x) + 2\cos(x))\right))/(\mathrm dx) = e^(3x)(\sin(x)+7\cos(x))}`