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Rewrite the function.

April 26, 2026Computer Science

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Answer

Y' = 4\sec^2(2x)\tan(2x)

4 steps

To find $Y'$ , we need to differentiate $Y = \sec^2(2x)$ with respect to $x$ . We will use the chain rule.

Rewrite the function.
$Y = (\sec(2x))^2$
Apply the power rule and the chain rule.
The derivative of $f(u)^n$ is $n f(u)^{n-1} f'(u)$ . Here, $f(u) = \sec(2x)$ and $n=2$ .
$Y' = 2(\sec(2x))^{2-1} \cdot \frac{d}{dx}(\sec(2x))$ $Y' = 2\sec(2x) \cdot \frac{d}{dx}(\sec(2x))$
Differentiate $\sec(2x)$ using the chain rule.
The derivative of $\sec(u)$ is $\sec(u)\tan(u) \cdot u'$ . Here, $u = 2x$ .
$\frac{d}{dx}(\sec(2x)) = \sec(2x)\tan(2x) \cdot \frac{d}{dx}(2x)$ $\frac{d}{dx}(2x) = 2$
So,
$\frac{d}{dx}(\sec(2x)) = \sec(2x)\tan(2x) \cdot 2$
Substitute this back into the expression for $Y'$ .
$Y' = 2\sec(2x) \cdot (2\sec(2x)\tan(2x))$ $Y' = 4\sec^2(2x)\tan(2x)$