find the derivative of the function.\n$y = \\sqrt{\\frac{x}{x + 5}}$\n$y=$

Explanation:

Step1: Rewrite the function

First, rewrite $y = \sqrt{\frac{x}{x + 5}}$ as $y=\left(\frac{x}{x + 5}\right)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=\frac{x}{x + 5}$, then $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. First, find $u^\prime$ using the quotient - rule. The quotient - rule for $u=\frac{f(x)}{g(x)}$ where $f(x)=x$ and $g(x)=x + 5$ is $u^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}$. Here, $f^\prime(x)=1$ and $g^\prime(x)=1$. So $u^\prime=\frac{1\cdot(x + 5)-x\cdot1}{(x + 5)^2}=\frac{x + 5-x}{(x + 5)^2}=\frac{5}{(x + 5)^2}$.

Step3: Substitute $u$ and $u^\prime$ into the chain - rule formula

Since $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$ and $u=\frac{x}{x + 5}$, we have $y^\prime=\frac{1}{2}\left(\frac{x}{x + 5}\right)^{-\frac{1}{2}}\cdot\frac{5}{(x + 5)^2}$. Simplify the expression: [ \begin{align*} y^\prime&=\frac{1}{2}\cdot\frac{\sqrt{x + 5}}{\sqrt{x}}\cdot\frac{5}{(x + 5)^2}\ &=\frac{5}{2\sqrt{x}(x + 5)^{\frac{3}{2}}} \end{align*} ]

Answer:

$\frac{5}{2\sqrt{x}(x + 5)^{\frac{3}{2}}}$