7. - / 1 points find the derivative of the function. y = \\sqrt{\\frac{x}{x + 3}} y =

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{\frac{x}{x + 3}}$ as $y=\left(\frac{x}{x + 3}\right)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=\frac{x}{x + 3}$, 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 $g(x)=x + 3$) is $u^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}$. Since $f^\prime(x)=1$ and $g^\prime(x)=1$, we have $u^\prime=\frac{1\cdot(x + 3)-x\cdot1}{(x + 3)^2}=\frac{x + 3-x}{(x + 3)^2}=\frac{3}{(x + 3)^2}$.

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

$y^\prime=\frac{1}{2}\left(\frac{x}{x + 3}\right)^{-\frac{1}{2}}\cdot\frac{3}{(x + 3)^2}=\frac{3}{2\sqrt{\frac{x}{x + 3}}(x + 3)^2}=\frac{3}{2\sqrt{x}(x + 3)^{\frac{3}{2}}}$.

Answer:

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