Question

differentiate.
y = \frac{\sqrt{x}}{\sqrt{x}+5}
y =

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sqrt{x}=x^{\frac{1}{2}}$, $v=\sqrt{x}+5=x^{\frac{1}{2}} + 5$. First, find $u'$ and $v'$.
$u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v'=\frac{1}{2}x^{-\frac{1}{2}}$.

Step2: Substitute into quotient - rule formula

$y'=\frac{\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x}+5)-\sqrt{x}\cdot\frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x}+5)^{2}}$.

Step3: Simplify the numerator

$\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x}+5)-\sqrt{x}\cdot\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}+\frac{5}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\cdot\sqrt{x}=\frac{5}{2\sqrt{x}}$.
So, $y'=\frac{\frac{5}{2\sqrt{x}}}{(\sqrt{x}+5)^{2}}=\frac{5}{2\sqrt{x}(\sqrt{x}+5)^{2}}$.

Answer:

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