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 using the product rule and chain rule.
\( \displaystyle \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x} \sin x) + \frac{d}{dx} (\sin \sqrt{x}) \\ \displaystyle = \left( \frac{d}{dx}(\sqrt{x}) \cdot \sin x + \sqrt{x} \cdot \frac{d}{dx}(\sin x) \right) + \frac{d (\sin \sqrt{x})}{d(\sqrt{x})} \cdot \frac{d}{dx} (\sqrt{x}) \\ \displaystyle = \left( \frac{1}{2\sqrt{x}} \cdot \sin x + \sqrt{x} \cos x \right) + \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \\ \displaystyle = \left( \frac{\sin x}{2\sqrt{x}} + \sqrt{x} \cos x \right) + \frac{\cos \sqrt{x}}{2\sqrt{x}} \\ \displaystyle = \frac{\sin x + \cos \sqrt{x}}{2\sqrt{x}} + \sqrt{x} \cos x \\ \)Therefore, \(\displaystyle \boxed{\frac{dy}{dx} = \frac{\sin x + \cos \sqrt{x}}{2\sqrt{x}} + \sqrt{x} \cos x } \)
Please let me know in the comments if you find any error in this solution.