How do you differentiate #f(x)=cos5x * cot3x# using the product rule?

Question

How do you differentiate #f(x)=cos5x  *  cot3x# using the product rule?

Scarlett Allison · Accepted Answer

#-5sin5xcot3x-3csc^2(3x)cos5x# The derivative of a product is stated as follows: #color(blue)((u(x)*v(x))'=u'(x)*v(x)+v'(x)*u(x))# Take #u(x)=cos(5x)# and #v(x)=cot(3x)# Let's find #u'(x)# and #v'(x)# Knowing the derivative of trigonometric function that says: #(cosy)'=-y'siny# and #(cot(y))' =-y'(csc^2y)# So, #u'(x)=(cos5x)'=-(5x)'sin5x=-5sin5x# #v'(x)=(cot3x)'=-(3x)'csc^2(3x)=-3csc^2(3x)# Thus, #color(blue)(f'(x)=(u(x)*v(x))')# Substituting #u'(x)# and #v'(x)# in the above property we have: #=-5sin5xcot3x-3csc^2(3x)cos5x#

Isabella Ackerman · Answer

undefined To differentiate the function $ f(x) = \cos(5x) \cdot \cot(3x) $ using the product rule, we first identify the two functions being multiplied: $ \cos(5x) $ and $ \cot(3x) $. Then, we 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.

Let $ u = \cos(5x) $ and $ v = \cot(3x) $.

Now, we find the derivatives of $ u $ and $ v $:
- $ \frac{du}{dx} = -5\sin(5x) $
- $ \frac{dv}{dx} = -3\csc^2(3x) $

Using the product rule, we have:
$$ \frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + \frac{du}{dx} \cdot v $$
$$ = \cos(5x) \cdot (-3\csc^2(3x)) + (-5\sin(5x)) \cdot \cot(3x) $$
$$ = -3\cos(5x)\csc^2(3x) - 5\sin(5x)\cot(3x) $$

Therefore, the derivative of $ f(x) = \cos(5x) \cdot \cot(3x) $ with respect to $ x $ using the product rule is $ -3\cos(5x)\csc^2(3x) - 5\sin(5x)\cot(3x) $.