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

Question

William Abdalla · Accepted Answer

#-2/3 cos(x)cos(3x)+2sin(x)sin(3x)#

The following guidelines must be followed in order to answer this question:

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)#.

Charles Arocha · Answer

undefined To differentiate the function $ f(x) = \cos(3x) \left(-\frac{2}{3}\sin(x)\right) $ using the product rule:

Let $ u(x) = \cos(3x) $ and $ v(x) = -\frac{2}{3}\sin(x) $.

Then, $ u'(x) = -3\sin(3x) $ and $ v'(x) = -\frac{2}{3}\cos(x) $.

Applying the product rule:

$$ f'(x) = u(x)v'(x) + v(x)u'(x) $$

$$ = (\cos(3x))\left(-\frac{2}{3}\cos(x)\right) + \left(-\frac{2}{3}\sin(x)\right)(-3\sin(3x)) $$

$$ = -\frac{2}{3}\cos(3x)\cos(x) + 2\sin(x)\sin(3x) $$