Differentiate the function.
\[
\begin{array}{r}
y=\frac{\sqrt{x}}{3+x} \\
y^{\prime}=\square
\end{array}
\]
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Step 1 ：Given the function \(y=\frac{\sqrt{x}}{3+x}\), we need to find its derivative \(y'\).

Step 2 ：We use the quotient rule for differentiation, which states that the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

Step 3 ：First, we find the derivatives of the numerator and the denominator. The derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\) and the derivative of \(3+x\) is 1.

Step 4 ：Substituting these values into the quotient rule, we get \(y' = -\frac{\sqrt{x}}{(x + 3)^2} + \frac{1}{2\sqrt{x}(x + 3)}\).

Step 5 ：\(\boxed{y' = -\frac{\sqrt{x}}{(x + 3)^2} + \frac{1}{2\sqrt{x}(x + 3)}}\) is the final answer.