Question

differentiate the function.
$y = \frac{\sqrt{x}}{7 + x}$
$y=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=7 + x$.

Step2: Find $u'$ and $v'$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $u = x^{\frac{1}{2}}$, $u'=\frac{1}{2}x^{-\frac{1}{2}}$. And for $v = 7+x$, $v' = 1$.

Step3: Apply the quotient - rule

$y'=\frac{\frac{1}{2}x^{-\frac{1}{2}}(7 + x)-x^{\frac{1}{2}}\times1}{(7 + x)^{2}}$.

Step4: Simplify the numerator

$\frac{1}{2}x^{-\frac{1}{2}}(7 + x)-x^{\frac{1}{2}}=\frac{7}{2}x^{-\frac{1}{2}}+\frac{1}{2}x^{\frac{1}{2}}-x^{\frac{1}{2}}=\frac{7}{2\sqrt{x}}-\frac{1}{2}\sqrt{x}=\frac{7 - x}{2\sqrt{x}}$.

Step5: Write the final derivative

$y'=\frac{7 - x}{2\sqrt{x}(7 + x)^{2}}$.

Answer:

$\frac{7 - x}{2\sqrt{x}(7 + x)^{2}}$