The main tool you have at your disposal for finding the derivative of this function is the chain rule, which tells you that you can differentiate a function that depends on a variable u, which in turn depends on another variable x, by
color(blue)(d/dx(y) = d/(du)(y) * d/dx(u)
You're also going to use the power rule, which allows you to calculate the derivative of a variable x raised to the power a by
color(blue)(d/dx(x^a) = a * x^(a-1))
Other derivatives that will come in handy
d/dx(sinx) = cosx
and
d/dx(cosx) = -sinx
So, looking at your initial function
y = sin^2(cos(3x))
you can predict that you're going to use the power rule once and the chain rule twice.
Use the chain rule to express the derivative of sin^2(cos(3x)) in terms of u = cos(3x)
d/dx(y) = d/(du)sin^2u * d/dx(u) " "color(purple)((1))
Now focus on d/(du)(sin^2u). You're going to have to use the chain rule again, only this time for
sinu = u_1, which implies that
sin^2u = u_1^2
This is equivalent to
d/(du)(sin^2u) = underbrace(d/(du_1) * u_1^2)_(color(blue)("power rule")) * d/(du)u_1
d/(du)(sin^2u) = 2u_1^(2-1) * d/(du)(sinu)
d/(du)(sin^2u) = 2sinu * cosu
Now go back to equation color(purple)((1))
y^' = 2sinu * cosu * d/dx(u)
Next, focus on d/dx(u) = d/dx(cos(3x)). Use the chain rule again, but this time use cos(u_2) and u_2 = 3x to get
d/dx(cos(3x)) = d/(du_2)cos(u_2) * d/(dx)(u_2)
d/dx(cos(3x)) = -sinu_2 * 3d/dx(x)
d/dx(cos(3x)) = -sin(3x) * 3
Take this back to equation color(purple((1)) to get
y^' = 2sinu * cosu * (-3 * sin(3x))
Remember that u = cos(3x), which means that you have
y^' = -3 * 2 * sin(cos(3x)) * cos(cos(3x)) * sin(3x)
y^' = color(green)(-6 * sin(3x) * sin(cos(3x)) * cos(cos(3x)))
