Question

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

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)