Question

find the derivative of $\frac{9}{3x^{2}-2x}$

Explanation:

Step1: Identify the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 9$, $v=3x^{2}-2x$.

Step2: Find the derivatives of $u$ and $v$

The derivative of a constant $u = 9$ is $u^\prime=0$. The derivative of $v = 3x^{2}-2x$ using the power - rule ($(x^{n})^\prime=nx^{n - 1}$) is $v^\prime=(3x^{2}-2x)^\prime=6x - 2$.

Step3: Apply the quotient - rule

Substitute $u = 9$, $u^\prime=0$, $v = 3x^{2}-2x$, and $v^\prime=6x - 2$ into the quotient - rule formula:
\[

$$\begin{align*} y^\prime&=\frac{0\times(3x^{2}-2x)-9\times(6x - 2)}{(3x^{2}-2x)^{2}}\\ &=\frac{-9(6x - 2)}{(3x^{2}-2x)^{2}}\\ &=\frac{-54x + 18}{(3x^{2}-2x)^{2}} \end{align*}$$

\]

Answer:

$\frac{-54x + 18}{(3x^{2}-2x)^{2}}$