Question

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

Answer 1

`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))`