To find \( \frac{dy}{dx} \) for the function \( y = x \sin(x) \), we need to apply the product rule of differentiation.

The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x) v(x) \) with respect to \( x \) is given by:

\( \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \).

Let \( u(x) = x \) and \( v(x) = \sin(x) \). Then:

\( u'(x) = 1 \) (since the derivative of \( x \) with respect to \( x \) is 1) and \( v'(x) = \cos(x) \) (since the derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \)).

Now apply the product rule:

\( \frac{dy}{dx} = u'(x) v(x) + u(x) v'(x) \),

\( \frac{dy}{dx} = (1)(\sin(x)) + (x)(\cos(x)) \),

\( \frac{dy}{dx} = \sin(x) + x \cos(x) \).

Now, when \( x = \frac{\pi}{2} \), we have:

\( \frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) \),

\( \frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = 1 + \frac{\pi}{2} \cdot 0 \),

\( \frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = 1 \).

So, the correct answer is 1