To differentiate the function $y = 2\cos\left(\frac{1}{2}x\right)$ and then substitute $x = \frac{\pi}{2}$, follow these steps:

Step 1: Differentiate the Function

We will use the chain rule to differentiate $y = 2\cos\left(\frac{1}{2}x\right)$.

Let $u = \frac{1}{2}x$, so the function becomes: $y = 2\cos(u)$

Now, differentiate with respect to $x$:

$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}[\cos(u)] \cdot \frac{du}{dx}$

The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$, and the derivative of $u = \frac{1}{2}x$ with respect to $x$ is $\frac{1}{2}$.

So, we have:

$\frac{dy}{dx} = 2 \cdot (-\sin(u)) \cdot \frac{1}{2}$

Simplify:

$\frac{dy}{dx} = -\sin\left(\frac{1}{2}x\right)$

Step 2: Substitute $x = \frac{\pi}{2}$ into the Derivative

Now, substitute $x = \frac{\pi}{2}$ into the derivative:

$\frac{dy}{dx}\bigg|_{x = \frac{\pi}{2}} = -\sin\left(\frac{1}{2} \cdot \frac{\pi}{2}\right)$

Simplify the argument of the sine function:

$\frac{dy}{dx}\bigg|_{x = \frac{\pi}{2}} = -\sin\left(\frac{\pi}{4}\right)$

We know that $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, so:

$\frac{dy}{dx}\bigg|_{x = \frac{\pi}{2}} = -\frac{\sqrt{2}}{2}$

Thus, the value of the derivative at $x = \frac{\pi}{2}$ is $-\frac{\sqrt{2}}{2}$.

Would you like further explanation or have any questions?

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Here are 5 related questions to explore further:

1. How do you differentiate trigonometric functions involving multiple angles?
2. What is the chain rule, and how does it apply to composite functions?
3. How do you evaluate trigonometric functions at special angles?
4. What is the significance of the derivative in determining the slope of a curve?
5. How would you differentiate a function like $y = \sin^2(x)$ or $y = \cos^3(x)$?

Tip: When working with trigonometric functions, it's helpful to remember the values of sine and cosine at key angles like $\frac{\pi}{4}$, $\frac{\pi}{2}$, and $\pi$.