How do you take the derivative of #sec^-1 (3x^2)#?

For your function #y = sec^(-1)(3x^2)#, you know that you can write

#sec(y) = 3x^2#

Use implicit differentiation and the fact that

#color(blue)(d/dx(secx) = secx * tanx)#

to write

#d/(dy)(secy) * (dy)/dx = d/dx(3x^2)#

#secy * tany * (dy)/dx = 6x#

Isolate #(dy)/dx# on one side of the equation to get

#(dy)/dx = (6x)/(secy * tany)#

Now, #"arcsec"(x)# has a range of #[0, pi/2) uu (pi/2, pi]#. If you evaluate the product of #secx# and #tanx# over the first interval, #[0, pi/2)#, you will always end up with a positive number because both the value of the tangent, and the value of secant are positive in that interval.

For the second interval, #(pi/2, 0]#, the product will once again be positive, since the tangent and the secant are both negative in that interval.

The #secx * tanx# product will always be positive, which means that you can write

#secx * tanx = sqrt( (secx * tanx)^2)#

Take this back to your derivative calculation

#(dy)/dx = (6x)/sqrt( (secy * tany)^2) = (6x)/(sqrt(sec^2y * tan^2y))#

Another important thing to remember is that

#color(blue)(sec^2x = 1 + tan^2x#

which means that you have

#(dy)/dx = (6x)/sqrt(sec^2y * (sec^2y - 1))#

#(dy)/dx = (6x)/(|sec^2y| * sqrt(sec^2y-1))#

Since #secy = 3x^2#, you get that

#(dy)/dx = (6x)/(underbrace(3x^2)_(color(red)("always positive")) * sqrt((3x^2)^2 - 1))#

Finally, the derivative of #y = sec^(-1)(3x^2)# is equal to

#(dy)/dx = color(green)(2/(x * sqrt(9x^4-1)))#