Question

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

Explanation:

Step1: Identify u and v

Let $u = 9$ and $v=3x^{2}-2x$.

Step2: Find u' and v'

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

Step3: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Substituting $u = 9$, $u' = 0$, $v=3x^{2}-2x$, and $v'=6x - 2$ into the quotient - rule formula, we get $y'=\frac{0\times(3x^{2}-2x)-9\times(6x - 2)}{(3x^{2}-2x)^{2}}$.

Step4: Simplify the expression

$y'=\frac{-9(6x - 2)}{(3x^{2}-2x)^{2}}=\frac{-54x + 18}{(3x^{2}-2x)^{2}}$.

Answer:

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