Find second order derivative of `e^(6x)cos3x`

Explanation

Let `y=e^(6x)" cos"3x`
`implies(dy)/(dx)=(d)/(dx)(e^(6x)"cos"3x)`
`=e^(6x)*(-3" sin"3x)+"cos"3x*(6e^(6x))`
`=e^(6x)(-3" sin"3x+6" cos"3x)`
`implies (d^(2)y)/(dx^(2))=(d)/(dx)[e^(6x)(-3" sin"3x+6" cos"3x)]`
`=e^(6x)(-9" cos"3x-18" sin"3x)+(-3" sin"3x+6" cos"3x)*(6e^(6x))`
`=e^(6x)(-9" cos"3x-18" sin"3x-18" sin"3x+36" cos"3x)`
`=e^(6x)(27" cos"3x-36" sin"3x)`
`=9e^(6x)(3" cos"3x-4" sin"3x)`