1. **State the problem:** We are given two functions $y_1 = \sin(x^2)$ and $y_2 = x \cos(x^2)$ and asked to evaluate their Wronskian $W(y_1,y_2)$ at $x = \sqrt{\pi}$. 2. **Recall the Wronskian formula:** For two functions $y_1$ and $y_2$, the Wronskian is defined as $$W(y_1,y_2) = y_1 y_2' - y_2 y_1'$$ where $y_1'$ and $y_2'$ are the derivatives of $y_1$ and $y_2$ with respect to $x$. 3. **Find the derivatives:** - $y_1 = \sin(x^2)$ Using the chain rule, $$y_1' = \frac{d}{dx} \sin(x^2) = \cos(x^2) \cdot 2x = 2x \cos(x^2)$$ - $y_2 = x \cos(x^2)$ Using the product rule, $$y_2' = \frac{d}{dx} (x \cos(x^2)) = 1 \cdot \cos(x^2) + x \cdot \frac{d}{dx} \cos(x^2)$$ Derivative of $\cos(x^2)$ is $-\sin(x^2) \cdot 2x = -2x \sin(x^2)$, so $$y_2' = \cos(x^2) - 2x^2 \sin(x^2)$$ 4. **Substitute into the Wronskian formula:** $$W = y_1 y_2' - y_2 y_1' = \sin(x^2) \left( \cos(x^2) - 2x^2 \sin(x^2) \right) - x \cos(x^2) \cdot 2x \cos(x^2)$$ Simplify terms: $$W = \sin(x^2) \cos(x^2) - 2x^2 \sin^2(x^2) - 2x^2 \cos^2(x^2)$$ 5. **Evaluate at $x = \sqrt{\pi}$:** Calculate $x^2 = (\sqrt{\pi})^2 = \pi$. Substitute: $$W(\sqrt{\pi}) = \sin(\pi) \cos(\pi) - 2\pi \sin^2(\pi) - 2\pi \cos^2(\pi)$$ Recall $\sin(\pi) = 0$ and $\cos(\pi) = -1$, so $$W(\sqrt{\pi}) = 0 \cdot (-1) - 2\pi \cdot 0 - 2\pi \cdot (-1)^2 = 0 - 0 - 2\pi = -2\pi$$ **Final answer:** $$\boxed{W(\sqrt{\pi}) = -2\pi}$$