Question

the function $y = \frac{9x^2 + 3x + 3}{\sqrt{x}}$ 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. $y = \frac{9x^2 + 3x + 3}{\sqrt{x}} = \frac{9x^2}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{3}{\sqrt{x}}$ remembering that $\sqrt{x} = x^{1/2}$ and that $\frac{x^n}{x^m} = x^{n - m}$, we will get the following. $\frac{9x^2}{\sqrt{x}} = 9x^{\square}$ $\frac{3x}{\sqrt{x}} = 3x^{\square}$ $\frac{3}{\sqrt{x}} = 3x^{\square}$

Explanation:

Step1: Simplify $\frac{9x^2}{\sqrt{x}}$

$\frac{9x^2}{x^{1/2}} = 9x^{2 - 1/2} = 9x^{3/2}$

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

$\frac{3x}{x^{1/2}} = 3x^{1 - 1/2} = 3x^{1/2}$

Step3: Simplify $\frac{3}{\sqrt{x}}$

$\frac{3}{x^{1/2}} = 3x^{-1/2}$

Answer:

First blank: $3/2$
Second blank: $1/2$
Third blank: $-1/2$