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

1 Answer
Feb 27, 2017

Answer:

#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#