The function #y = sec^2(2x)# can be rewritten as #y = sec(2x)^2# or #y = g(x)^2# which should clue us in as a good candidate for the power rule.
The power rule: #dy/dx = n* g(x)^(n-1) * d/dx(g(x))#
where #g(x) = sec(2x)# and #n=2# in our example.
Plugging these values into the power rule gives us
#dy/dx = 2 * sec(2x) ^ 1 *d/dx(g(x))#
Our only unknown remains #d/dx(g(x))#.
To find the derivative of #g(x) = sec(2x)#, we need to use the chain rule because the inner part of #g(x)# is actually another function of #x#. In other words, #g(x) = sec(h(x))#.
The chain rule: #g(h(x))' = g'(h(x)) * h'(x)# where
#g(x) = sec(h(x))# and
#h(x) = 2x#
#g'(h(x)) = sec(h(x))tan(h(x))#
#h'(x) = 2#
Let's use all of these values in the chain rule formula:
#d/dx(g(x)) = d/dx(g(h(x))) = sec(2x)tan(x) * 2 = 2sec(2x)tan(x)#
Now we can finally plug back this result into the power rule.
#dy/dx = 2 * sec(2x) ^ 1 * d/dx(g(x))#
#dy/dx = 2sec(2x) * 2sec(2x)tan(x) = 4sec^2(2x)tan(2x)#
