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

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Answer 1 · 2015-05-14T12:07:50

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:

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

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