Step 1: Problem Identification:
The objective is to compute the derivative of a nested function, necessitating the iterative application of the chain rule. Subsequently, the derived expression will be evaluated at a given point.
Step 2: Governing Principle:
The chain rule dictates that for \( y = f(g(h(x))) \), the derivative is: \[ \frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \] The function is decomposed as \( y = \sin(u) \), \( u = \cos(v) \), and \( v = x^2 \).
Step 3: Derivation and Evaluation:
Given the function \( y = \sin(\cos(x^2)) \), we apply the chain rule for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{d}{dx} \sin(\cos(x^2)) \] \[ = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)) \] \[ = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot \frac{d}{dx}(x^2) \] \[ = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) \] \[ \frac{dy}{dx} = -2x \sin(x^2) \cos(\cos(x^2)) \] Evaluate this derivative at \( x = \frac{\sqrt{\pi}}{2} \). First, calculate the intermediate terms:
\[ x^2 = \left(\frac{\sqrt{\pi}}{2}\right)^2 = \frac{\pi}{4} \] Thus:
\[ \sin(x^2) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] \[ \cos(x^2) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] Substitute these values into the derivative formula:
\[ \frac{dy}{dx} \bigg|_{x=\frac{\sqrt{\pi}}{2}} = -2\left(\frac{\sqrt{\pi}}{2}\right) \cdot \sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\cos\left(\frac{\pi}{4}\right)\right) \] \[ = -\sqrt{\pi} \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot \cos\left(\frac{1}{\sqrt{2}}\right) \] \[ = -\frac{\sqrt{\pi}}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \] This result is equivalent to \( -\sqrt{\frac{\pi}{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \).
Step 4: Conclusive Result:
The derivative of the function evaluated at \( x = \frac{\sqrt{\pi}}{2} \) is \( -\frac{\sqrt{\pi}}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \). This corresponds to option (A).