Find $y^{\prime}$ and $y^{\prime \prime}$. \[ y=\sqrt{x} \ln (x) \] \[ y^{\prime}= \] \[ y^{\prime \prime}= \]

Step 1 ：Let \(u = \sqrt{x}\) and \(v = \ln(x)\).

Step 2 ：Then, \(u' = \frac{1}{2\sqrt{x}}\) and \(v' = \frac{1}{x}\).

Step 3 ：Applying the product rule, we get \(y' = u'v + uv'\).

Step 4 ：\(y' = \frac{1}{2\sqrt{x}}\ln(x) + \sqrt{x}\frac{1}{x}\).

Step 5 ：Simplify to get \(y' = \frac{\ln(x) + 2}{2\sqrt{x}}\).

Step 6 ：Now, to find the second derivative \(y''\), we will need to use the quotient rule.

Step 7 ：Let \(p = \ln(x) + 2\) and \(q = 2\sqrt{x}\).

Step 8 ：Then, \(p' = \frac{1}{x}\) and \(q' = \frac{1}{\sqrt{x}}\).

Step 9 ：Applying the quotient rule, we get \(y'' = \frac{p'q - pq'}{q^2}\).

Step 10 ：\(y'' = \frac{\frac{1}{x}2\sqrt{x} - (\ln(x) + 2)\frac{1}{\sqrt{x}}}{(2\sqrt{x})^2}\).

Step 11 ：Simplify to get \(y'' = \frac{2(\sqrt{x} - 1) - \ln(x)}{4x}\).

Step 12 ：So, the first derivative \(y'\) is \(\boxed{\frac{\ln(x) + 2}{2\sqrt{x}}}\) and the second derivative \(y''\) is \(\boxed{\frac{2(\sqrt{x} - 1) - \ln(x)}{4x}}\).