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

1 Answer
May 14, 2015

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 product rule, 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)#