How do you differentiate #f(x)=cos(3x)*(cosx)# using the product rule?

Question

Emily Ammerman · Accepted Answer

#-cos(3x)sin(x) - 3sin(3x)cos(x)# Use the product rule #d/dx (f(x)g(x)) =f(x)g'(x) + f'(x)g(x)# #f(x) = cos(3x)# so # f'(x) = -3sin(3x)# #g(x) =cos(x)# so #g'(x) = -sin(x)# Using the produce rule gives #f'(x) = cos(3x)*(-sin(x) +(-3sin(3x)*cos(x)# #=-cos(3x)sin(x) - 3sin(3x)cos(x)#

Paisley Johnson · Answer

undefined To differentiate $ f(x) = \cos(3x) \cdot \cos(x) $ using the product rule, you would follow these steps:

1. Identify the functions $ u(x) = \cos(3x) $ and $ v(x) = \cos(x) $.
2. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Differentiate $ u(x) $ and $ v(x) $ separately.
4. Substitute the derivatives and the original functions into the product rule formula.
5. Simplify the expression to get the derivative of $ f(x) $.

Paisley Johnson · Answer

undefined To differentiate $ f(x) = \cos(3x) \cdot \cos(x) $ using the product rule, follow these steps:

1. Identify the two functions being multiplied: $ \cos(3x) $ and $ \cos(x) $.
2. Apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
3. Differentiate each function separately.
4. Substitute the derivatives into the product rule formula.

Let's denote $ u = \cos(3x) $ and $ v = \cos(x) $.

The derivative of $ u $ with respect to $ x $, denoted as $ u' $, is $ -3\sin(3x) $, and the derivative of $ v $ with respect to $ x $, denoted as $ v' $, is $ -\sin(x) $.

Now, apply the product rule:

$$ f'(x) = u'v + uv' $$
$$ f'(x) = (-3\sin(3x))\cdot(\cos(x)) + (\cos(3x))\cdot(-\sin(x)) $$

So, the derivative of $ f(x) = \cos(3x) \cdot \cos(x) $ using the product rule is:

$$ f'(x) = -3\sin(3x) \cdot \cos(x) + \cos(3x) \cdot (-\sin(x)) $$

$$ f'(x) = -3\sin(3x)\cos(x) - \sin(x)\cos(3x) $$