Question

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

Answer 1

`g'(x) = 3sin6xcos3x + 6sin3xcos6x`

Explanation:

The product rule says that if

`f(x) = a(x) * b(x)`

then

`f'(x) = a'(x)b(x) + b'(x)a(x)`

In the above example,

`g(x) = sin(3x)sin(6x)`,

so

`a(x) = sin(3x), b(x)=sin(6x)`

Now we have our two separate functions, we can differentiate them. However, these functions have insides (`3x` or `6x`) and outsides (`sin`), so to differentiate them we multiply the derivative of the inside by the derivative of the outside with the inside left the same. In other words,

`d/dx sin(ax + b) = acos(ax + b)`

so

`a(x)=sin(3x), b(x)=sin(6x)`

`a'(x)=3cos(3x), b'(x)=6cos(6x)`

Therefore the derivative of the whole thing, if you remember back to the product rule equation, is

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

`g'(x) = 3sin6xcos3x + 6sin3xcos6x`