let ( f(x)=\frac{x^{2}(x - 9)^{2}}{(x^{2}+8)^{9}} ). use logarithmic differentiation to determine the…

Explanation:

Step1: Take the natural logarithm of (f(x))

$$\ln f(x)=\ln\left(\frac{x^{2}(x - 9)^{2}}{(x^{2}+8)^{9}}\right)$$ Using the logarithm rules (\ln\frac{a}{b}=\ln a-\ln b) and (\ln(ab)=\ln a+\ln b) and (\ln a^{n}=n\ln a), we get: $$\ln f(x)=2\ln x + 2\ln(x - 9)-9\ln(x^{2}+8)$$

Step2: Differentiate both sides with respect to (x)

Differentiate the left - hand side: (\frac{d}{dx}(\ln f(x))=\frac{f^{\prime}(x)}{f(x)}) Differentiate the right - hand side: Using the chain rule ((\ln u)^\prime=\frac{u^\prime}{u}) For (y = 2\ln x), (y^\prime=\frac{2}{x}) For (y = 2\ln(x - 9)), (y^\prime=\frac{2}{x - 9}) For (y=-9\ln(x^{2}+8)), (y^\prime=-9\times\frac{2x}{x^{2}+8}=-\frac{18x}{x^{2}+8})

So (\frac{f^{\prime}(x)}{f(x)}=\frac{2}{x}+\frac{2}{x - 9}-\frac{18x}{x^{2}+8})

Step3: Solve for (f^{\prime}(x))

Since (f(x)=\frac{x^{2}(x - 9)^{2}}{(x^{2}+8)^{9}}), then [ \begin{align*} f^{\prime}(x)&=f(x)\left(\frac{2}{x}+\frac{2}{x - 9}-\frac{18x}{x^{2}+8}\right)\ &=\frac{x^{2}(x - 9)^{2}}{(x^{2}+8)^{9}}\left(\frac{2}{x}+\frac{2}{x - 9}-\frac{18x}{x^{2}+8}\right)\ &=\frac{x(x - 9)^{2}}{(x^{2}+8)^{9}}\times2+\frac{x^{2}(x - 9)}{(x^{2}+8)^{9}}\times2-\frac{18x^{3}(x - 9)^{2}}{(x^{2}+8)^{10}}\ &=\frac{2x(x - 9)^{2}}{(x^{2}+8)^{9}}+\frac{2x^{2}(x - 9)}{(x^{2}+8)^{9}}-\frac{18x^{3}(x - 9)^{2}}{(x^{2}+8)^{10}} \end{align*} ]

Answer:

(f^{\prime}(x)=\frac{2x(x - 9)^{2}}{(x^{2}+8)^{9}}+\frac{2x^{2}(x - 9)}{(x^{2}+8)^{9}}-\frac{18x^{3}(x - 9)^{2}}{(x^{2}+8)^{10}})