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

Jul 23, 2014

The function $y = {\sec}^{2} \left(2 x\right)$ can be rewritten as $y = \sec {\left(2 x\right)}^{2}$ or $y = g {\left(x\right)}^{2}$ which should clue us in as a good candidate for the power rule.

The power rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = n \cdot g {\left(x\right)}^{n - 1} \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

where $g \left(x\right) = \sec \left(2 x\right)$ and $n = 2$ in our example.

Plugging these values into the power rule gives us

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sec {\left(2 x\right)}^{1} \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

Our only unknown remains $\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$.

To find the derivative of $g \left(x\right) = \sec \left(2 x\right)$, we need to use the chain rule because the inner part of $g \left(x\right)$ is actually another function of $x$. In other words, $g \left(x\right) = \sec \left(h \left(x\right)\right)$.

The chain rule: $g \left(h \left(x\right)\right) ' = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$ where

$g \left(x\right) = \sec \left(h \left(x\right)\right)$ and

$h \left(x\right) = 2 x$

$g ' \left(h \left(x\right)\right) = \sec \left(h \left(x\right)\right) \tan \left(h \left(x\right)\right)$

$h ' \left(x\right) = 2$

Let's use all of these values in the chain rule formula:

$\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(g \left(h \left(x\right)\right)\right) = \sec \left(2 x\right) \tan \left(x\right) \cdot 2 = 2 \sec \left(2 x\right) \tan \left(x\right)$

Now we can finally plug back this result into the power rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cdot \sec {\left(2 x\right)}^{1} \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \sec \left(2 x\right) \cdot 2 \sec \left(2 x\right) \tan \left(x\right) = 4 {\sec}^{2} \left(2 x\right) \tan \left(2 x\right)$