use logarithmic differentiation to evaluate ( f^{prime}(x) ).\n\n( f(x)=\frac{(x + 1)^{12}}{(3 x-9)^{9}}…

Explanation:

Step1: Take the natural logarithm of both sides

$$\ln f(x)=\ln\frac{(x + 1)^{12}}{(3x-9)^{9}}$$ Using the property of logarithms (\ln\frac{a}{b}=\ln a-\ln b), we get: $$\ln f(x)=12\ln(x + 1)-9\ln(3x - 9)$$

Step2: Differentiate both sides with respect to (x)

The derivative of (\ln u) with respect to (x) is (\frac{u'}{u}). For (y = 12\ln(x + 1)-9\ln(3x - 9)), we have: The derivative of (12\ln(x + 1)) is (12\times\frac{1}{x + 1}\times(x + 1)') (by the chain - rule). Since ((x + 1)'=1), it is (\frac{12}{x + 1}). The derivative of (-9\ln(3x - 9)) is (-9\times\frac{1}{3x - 9}\times(3x - 9)'). Since ((3x - 9)'=3), it is (-9\times\frac{3}{3x - 9}=-\frac{9}{x - 3}). So, (\frac{f'(x)}{f(x)}=\frac{12}{x + 1}-\frac{9}{x - 3})

Step3: Solve for (f'(x))

Since (f(x)=\frac{(x + 1)^{12}}{(3x-9)^{9}}), then (f'(x)=f(x)\left(\frac{12}{x + 1}-\frac{9}{x - 3}\right)) [ \begin{align*} f'(x)&=\frac{(x + 1)^{12}}{(3x-9)^{9}}\left(\frac{12}{x + 1}-\frac{9}{x - 3}\right)\ &=\frac{(x + 1)^{12}}{(3x-9)^{9}}\times\frac{12(x - 3)-9(x + 1)}{(x + 1)(x - 3)}\ &=\frac{(x + 1)^{12}}{(3x-9)^{9}}\times\frac{12x-36-9x - 9}{(x + 1)(x - 3)}\ &=\frac{(x + 1)^{12}}{(3x-9)^{9}}\times\frac{3x-45}{(x + 1)(x - 3)}\ &=\frac{(x + 1)^{11}(3x - 45)}{(3x-9)^{9}(x - 3)} \end{align*} ] We can factor out a (3) from (3x-45) and (3x - 9): Since (3x-45 = 3(x - 15)) and (3x-9=3(x - 3)), then ((3x - 9)^{9}=3^{9}(x - 3)^{9}) [ \begin{align*} f'(x)&=\frac{(x + 1)^{11}\times3(x - 15)}{3^{9}(x - 3)^{9}(x - 3)}\ &=\frac{(x + 1)^{11}(x - 15)}{3^{8}(x - 3)^{10}}\ &=\frac{(x + 1)^{11}(x - 15)}{6561(x - 3)^{10}} \end{align*} ]

Answer:

(f'(x)=\frac{(x + 1)^{11}(x - 15)}{6561(x - 3)^{10}})