How do you find the second derivative of ##sec^2 (2x)##?

By definition, ##sec^2(u) = (1 + tan(u))##

Considering, in our case, ##u = 2x##, we can derive your function as follows:

##(dy)/(du) = 0 + u'*sec^2(u)##

Deriving a constant (number ##1##) equals zero, so all we have left is the derived of ##tan(u)##.

Now, in order to derive again, we must remember that the derivative of ##sec^2(u)## is ##u'*sec(u)*tan(u)##.

However, we have a product - the two factors are ##u'## and ##sec^2(u)##. By the , we must proceed as follows:

##(d^2y)/(du^2) = u''*sec^2(u) + u'*u'*sec(u)*tan(u)##

But we know that ##u=2x## and, consequently, ##u'=2## and ##u''=0##.

Substituting these terms related to ##u## in our second derivative, we'll have:

##(d^2y)/(dx^2) = 0*sec^2(2x) + 2*2*sec(2x)*tan(2x)##

Final answer, then, is:

##(d^2y)/(dx^2) = 4*sec(2x)*tan(2x)##