# What is the derivative of y=sec^2(2x)?

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.

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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 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)##

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