Question

the function $y = \frac{7x^{2}+3x + 7}{\sqrt{x}}$ is a quotient. it is important to remember that the derivative of a quotient $f(x)/g(x)$ is not the quotient of the derivatives $f(x)/g(x)$. instead, we find the derivative by first simplifying the quotient. we can re - write it as follows.
$y=\frac{7x^{2}+3x + 7}{\sqrt{x}}=\frac{7x^{2}}{\sqrt{x}}+\frac{3x}{\sqrt{x}}+\frac{7}{\sqrt{x}}$
remembering that $\sqrt{x}=x^{1/2}$ and that $\frac{x^{n}}{x^{m}}=x^{n - m}$, we will get the following.
$\frac{7x^{2}}{\sqrt{x}} = 7x^{square}$
$\frac{3x}{\sqrt{x}} = 3x^{square}$
$\frac{7}{\sqrt{x}} = 7x^{square}$

Explanation:

Step1: Simplify $\frac{7x^{2}}{\sqrt{x}}$

Using the rule $\frac{x^{n}}{x^{m}}=x^{n - m}$ and $\sqrt{x}=x^{\frac{1}{2}}$, we have $\frac{7x^{2}}{x^{\frac{1}{2}}}=7x^{2-\frac{1}{2}} = 7x^{\frac{4 - 1}{2}}=7x^{\frac{3}{2}}$.

Step2: Simplify $\frac{3x}{\sqrt{x}}$

Applying the same rule, $\frac{3x}{x^{\frac{1}{2}}}=3x^{1-\frac{1}{2}}=3x^{\frac{1}{2}}$.

Step3: Simplify $\frac{7}{\sqrt{x}}$

Using the rule, $\frac{7}{x^{\frac{1}{2}}}=7x^{-\frac{1}{2}}$.

Answer:

$\frac{7x^{2}}{\sqrt{x}} = 7x^{\frac{3}{2}}$, $\frac{3x}{\sqrt{x}}=3x^{\frac{1}{2}}$, $\frac{7}{\sqrt{x}} = 7x^{-\frac{1}{2}}$