given ( f(x) = \frac{4x^2}{x^2 - 6x} ), find ( f(x) ).

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Accepted Answer

# Explanation:
## Step1: Identify the function type
The function $ f(x)=\frac{4x^{2}}{x^{2}-6x} $ is a rational function, so we use the quotient rule for differentiation. The quotient rule states that if $ f(x)=\frac{u(x)}{v(x)} $, then $ f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{[v(x)]^{2}} $, where $ u(x) = 4x^{2} $ and $ v(x)=x^{2}-6x $.

## Step2: Find $ u^{\prime}(x) $ and $ v^{\prime}(x) $
For $ u(x) = 4x^{2} $, using the power rule $ \frac{d}{dx}(x^{n})=nx^{n - 1} $, we have $ u^{\prime}(x)=4\times2x=8x $.

For $ v(x)=x^{2}-6x $, using the power rule, $ v^{\prime}(x)=2x - 6 $.

## Step3: Apply the quotient rule
Substitute $ u(x) $, $ u^{\prime}(x) $, $ v(x) $, and $ v^{\prime}(x) $ into the quotient rule formula:

$$
\begin{align*}
f^{\prime}(x)&=\frac{(8x)(x^{2}-6x)-4x^{2}(2x - 6)}{(x^{2}-6x)^{2}}\
&=\frac{8x^{3}-48x^{2}-8x^{3}+24x^{2}}{(x^{2}-6x)^{2}}\
&=\frac{(8x^{3}-8x^{3})+(-48x^{2}+24x^{2})}{(x^{2}-6x)^{2}}\
&=\frac{-24x^{2}}{(x^{2}-6x)^{2}}\
&=\frac{-24x^{2}}{x^{2}(x - 6)^{2}} \quad (\text{Factor }x^{2}\text{ from denominator})\
&=\frac{-24}{(x - 6)^{2}} \quad (\text{Cancel }x^{2}\text{, }x
eq0\text{ and }x
eq6)
\end{align*}
$$

# Answer:
$ f^{\prime}(x)=\frac{-24}{(x - 6)^{2}} $ (or equivalent simplified form)

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the function type

Step2: Find \( u^{\prime}(x) \) and \( v^{\prime}(x) \)

Step3: Apply the quotient rule

Answer: