1. **State the problem:** We need to find the derivative of the composition function $(f \circ g)(x) = f(g(x))$ evaluated at $x = \sqrt{\pi}$ given $f(x) = \sin(x)$ and $g(x) = x^{2}$. 2. **Find $(f \circ g)(x)$:** $$(f \circ g)(x) = f(g(x)) = \sin(x^{2})$$ 3. **Compute the derivative using the chain rule:** The derivative of $f(g(x))$ is $$ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) $$ We know: $$ f'(x) = \cos(x), \quad g'(x) = 2x $$ So, $$ (f \circ g)'(x) = \cos(x^{2}) \cdot 2x $$ 4. **Evaluate at $x = \sqrt{\pi}$:** $$ (f \circ g)'(\sqrt{\pi}) = \cos((\sqrt{\pi})^{2}) \cdot 2 \sqrt{\pi} = \cos(\pi) \cdot 2 \sqrt{\pi} = (-1) \cdot 2 \sqrt{\pi} = -2 \sqrt{\pi} $$ 5. **Final answer:** Option (d) $-2 \sqrt{\pi}$