Differentiate in terms of x:

Question

1] y = cos^3 5x

2] y = sec^2 3x

Answer 1

start out this way
3 [cos^2(5x)][-sin(5x)](5)

I do not have derivative of sec in my head so:
2 [sec (3x)][d/dx(1/cos(3x)](3)

and
d/dx[1/cos(3x)] = [sin(3x)(3)]/cos^2(3x)

then combine

Answer 2

1] let u (x) = cos(5x)
y = u^3

dy/dx = dy/du*du/dx = 3 u^2*d(cos 5x)/dx
= 3 cos^2(5x)*[-sin(5x)]*5
= -15 cos^(5x)* sin(5x)

I used the "chain rule" a second time to get d(cos 5x)/dx

2] Proceed similarly. Use the chain rule twice.
Let u(x) = sec(3x)
f(x) = u^2
df/dx = df/du * du/dx
= 2 u * d/dx (sec(3x))
Let v = 3x dv/dx = 3
df/dx = 2 u (d/dv)(sec(v))*(dv/dx)
= 6 sec(3x)*d/dv(sec v)
= 6 sec(3x)*tan(3x)*sec*(3x)
= 6 sec^2(3x)*tan(3x)

Answer 3

Where I wrote f and df/dx, you can substitute y and dy/dx