How do you find the second derivative of sec^2 (2x)sec2(2x)?

1 Answer
May 14, 2015

By definition, sec^2(u) = (1 + tan(u))sec2(u)=(1+tan(u))

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

(dy)/(du) = 0 + u'*sec^2(u)dydu=0+u'sec2(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)