The given function is \( \displaystyle y = \sqrt{x} \sin x + \sin(\sqrt{x}) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
We will differentiate each term separately using the Product Rule and the Chain Rule.
\( \begin{aligned} y &= \sqrt{x} \sin x + \sin(\sqrt{x}) \\ \frac{dy}{dx} &= \frac{d}{dx} (\sqrt{x} \sin x) + \frac{d}{dx} \sin(\sqrt{x}) \\ &= \frac{1}{2\sqrt{x}} \cdot \sin x + \sqrt{x} \cdot \cos x + \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x + \frac{\cos(\sqrt{x})}{2\sqrt{x}} \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x + \frac{\cos(\sqrt{x})}{2\sqrt{x}}} \)Please let me know in the comments if you find any errors in this solution.