What is the derivative of -sec^3(2x)

I'm not sure of what to do first.
I started like this: y'=-6(sec2x*tan2x)^2
But the answer is supposed to be y'=-6sec^3(2x)tan2x
What am i doing wrong? Thanks in advance!

Apply the chain rule one step at a time.

You have a function cubed.

y = -f(x)^3 where f(x) = sec(2x)

dy/dx = -3 f(x)^2 df/dx = -3 sec^2(2x) * d[sec(2x)]/dx

Now what's the derivative of sec(2x)? It's 2 sec(2x) tan(2x), applying the chain rule to the 2x.

So dy/dx = -3 sec^2(2x) * 2 sec(2x) tan(2x).

y = -sec^3(2x)
z = 2x
dz/dx = 2
t = sec(z)
dt/dz= sec(z)tan(z)
y = -t^3
dy/dt = -3t^2
dy/dx (what we want) = dz/dx * dt/dz * dy/dt
= 2 * sec(z) * sec(z)tan(z) * -3t^2
= 2* sec(2x) * sec(2x)tan(2x) * -3 sec^2(2x)
= -6sec^3(2x)tan(2x)

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d -sec³(2x)
—————  =  -3 sec²(2x) • sec(2x) tan(2x) • 2  =  -6sec³(2x) tan(2x)
       dx


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