Question

How do you differentiate `g(x) = (6x+9)sin(3x)` using the product rule?

Answer 1

`g'(x)=3(6x+9)cos(3x)+6sin(3x)`

Explanation:

First, we recall product rule :

`f(x)=a*b`
`f'(x)= a*db/dx +b*da/dx`

From the above question, we can think of `(6x+9)`as `a` and `sin(3x)` as `b`. So, we have the following:

`g'(x)= (6x+9)* d(sin(3x))/dx + sin(3x) *d(6x+9)/dx`
`=(6x+9)*cos(3x)d(3x)/dx+ sin(3x)*6`
`=3(6x+9)cos(3x)+6sin(3x)`