use logarithmic differentiation to evaluate ( f^{prime}(x) ).\n\n( f(x)=\frac{(x+1)^{11}}{(3 x-6)^{8}}…

Explanation:

Step1: Take natural logarithm on both sides

$$\ln f(x)=\ln\frac{(x + 1)^{11}}{(3x-6)^{8}}$$ Using the property of logarithms $\ln\frac{a}{b}=\ln a-\ln b$, we get: $$\ln f(x)=11\ln(x + 1)-8\ln(3x - 6)$$

Step2: Differentiate both sides with respect to (x)

Differentiate (y = 11\ln(x + 1)-8\ln(3x - 6)) using the chain rule ((\ln u)^\prime=\frac{u^\prime}{u}). For (y_1 = 11\ln(x + 1)), (y_1^\prime=\frac{11}{x + 1}\times(x + 1)^\prime=\frac{11}{x+1}) For (y_2=-8\ln(3x - 6)), (y_2^\prime=-8\times\frac{1}{3x - 6}\times(3x - 6)^\prime=-8\times\frac{3}{3x - 6}=-\frac{8}{x - 2}) So (\frac{f^\prime(x)}{f(x)}=\frac{11}{x + 1}-\frac{8}{x - 2})

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

Since (f(x)=\frac{(x + 1)^{11}}{(3x-6)^{8}}), then (f^\prime(x)=f(x)\left(\frac{11}{x + 1}-\frac{8}{x - 2}\right)) [ \begin{align*} f^\prime(x)&=\frac{(x + 1)^{11}}{(3x-6)^{8}}\left(\frac{11(x - 2)-8(x + 1)}{(x + 1)(x - 2)}\right)\ &=\frac{(x + 1)^{11}}{(3x-6)^{8}}\left(\frac{11x-22-8x - 8}{(x + 1)(x - 2)}\right)\ &=\frac{(x + 1)^{11}}{(3x-6)^{8}}\left(\frac{3x-30}{(x + 1)(x - 2)}\right)\ &=\frac{3(x + 1)^{10}(x - 10)}{(3x-6)^{8}(x - 2)} \end{align*} ]

Answer:

(f^\prime(x)=\frac{3(x + 1)^{10}(x - 10)}{(3x-6)^{8}(x - 2)})