How do you differentiate #f(x)=cos(3x)*(-2/3sinx)# using the product rule?

Here are some of the rules that need to be used to solve this question:

• Chain rule: #(df)/dx=(df)/(dg)*(dg)/dx#, where #f# and #g# are functions of #x#
• Product rule: #(d(u*v))/dx=u*(dv)/(dx)+v*(du)/(dx)#, where #u# and #v# are functions of #x#
• Constant factor rule (actually a special case of the product rule): #(d(ku))/dx=k (du)/dx#, where #k# is a constant and u is a function of #x#

First of all, we rewrite the original function to #-2/3*cos(3x)*sin(x)#. We will deal with the constant #-2/3# later and differentiate #cos(3x)*sin(x)# first (permissible by the constant factor rule).

We shall apply the product rule to find the derivative of #cos(3x)*sin(x)#. First, we call #u=cos(3x)# and #v=sin(x)#. Then, we need to find #u'# and #v'#. We know that #v'=cos(x)#. However, to solve #u'#, we need the chain rule.

The chain rule is used for finding composite functions. For the chain rule, we say that #f(x)=cos(x)# and #g(x)=3x#. The function is then #f(g(x))=cos(3x)#. Applying the chain rule, we find that the derivative of #cos(3x)# is #(d(3x))/dx*(d cos(3x))/(d(3x))=3*-sin(3x)=-3sin(3x)#.

Now, we have found #u'#. Substituting in the product rule, #cos(3x)*cos(x)-sin(x)*3sin(3x)=cos(x)cos(3x)-3sin(x)sin(3x)#.

Finally, we apply the constant factor rule, i.e. multiplying this function by #-2/3#, to obtain the final answer #-2/3 cos(x)cos(3x)+2sin(x)sin(3x)#.