How do you use the Product Rule to find the derivative of #sin(3x) cos(5x) #?

1 Answer
Aug 9, 2015

Answer:

#y^' = 3cos(3x) * cos(5x) -5sin(3x) * sin(5x)#

Explanation:

First, notice that you can write your function as a product of two functions, #f(x) = sin(3x)# and #g(x) = cos(5x)#

#y = f(x) * g(x)#

The product rule allows you to differentiate such functions by using the formula

#color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))#

In your case, you have

#d/dx(y) = [d/dx(sin(3x))] * cos(5x) + sin(3x) * d/dx(cos(5x))#

Now, in order to differentiate these two functions, you need to use the chain rule in the form #sinu_1# and #cosu_2#, with #u_1 = 3x# and #u_2 = 5x#.

This will get you

#d/dx(sinu_1) - [d/(du_1)sinu_1] * d/dx(u_1)#

#d/dx(sinu_1) = cosu_1 * d/dx(3x)#

#d/dx(sin(3x)) = cos(3x) * 3#

and

#d/dx(cosu_2) = [d/(du_2)cosu_2] * d/dx(u_2)#

#d/dx(cos(5x)) = -sin(5x) * 5#

Your target derivative will thus be

#y^' = [3 * cos(3x)] * cos(5x) + sin(3x) * [-5sin(5x)]#

#y^' = color(green)(3cos(3x) * cos(5x) -5sin(3x) * sin(5x))#