Rewrite the function.

Hey ELIAS, good to see you again. To find $Y'$, we need to differentiate $Y = \sec^2(2x)$ with respect to $x$. We will use the chain rule. Step 1: Rewrite the function. $$Y = (\sec(2x))^2$$ Step 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 $n=2$. $$Y' = 2(\sec(2x))^{2-1} \cdot \frac{d}{dx}(\sec(2x))$$ $$Y' = 2\sec(2x) \cdot \frac{d}{dx}(\sec(2x))$$ Step 3: 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, $$\frac{d}{dx}(\sec(2x)) = \sec(2x)\tan(2x) \cdot 2$$ Step 4: Substitute this back into the expression for $Y'$. $$Y' = 2\sec(2x) \cdot (2\sec(2x)\tan(2x))$$ $$Y' = 4\sec^2(2x)\tan(2x)$$ The derivative is: $$\boxed{Y' = 4\sec^2(2x)\tan(2x)}$$ Send me the next one 📸