Question

How do you differentiate `f(x)=e^(3x)cos2x` using the product rule?

Answer 1

`f' (x)=e^(3x)*[3*cos 2x-2*sin 2x]`

Explanation:

Given `f(x)=e^(3x)*cos(2x)`

Use the formula
`d/dx(uv)=u*d/dx(v)+v*d/dx(u)`
Let `u=e^(3x)` and `v=cos(2x)`

`d/dx(uv)=u*d/dx(v)+v*d/dx(u)`
`f' (x)=d/dx(e^(3x)*cos(2x))=e^(3x)*d/dx(cos(2x))+cos(2x)*d/dx(e^(3x))`

`f' (x)=e^(3x)*-sin(2x)*d/dx(2x)+cos(2x)*e^(3x)*d/dx(3x)`

`f' (x)=e^(3x)*(-sin(2x))(2)+cos(2x)e^(3x)(3)`

simplify by factoring common factors

`f' (x)=e^(3x)*[3*cos 2x-2*sin 2x]`

God bless...I hope the explanation is useful.