use logarithmic differentiation to evaluate ( f^{prime}(x) ).\n\n( f(x)=(4 x)^{ln 4 x} )\n\n( f^{prime}(x)=…

Explanation:

Step1: Take natural logarithm on both sides

$$\ln f(x)=\ln(4x)^{\ln 4x}$$ Using the property $\ln a^b = b\ln a$, we get $\ln f(x)=(\ln 4x)\cdot\ln(4x)=(\ln 4x)^2$.

Step2: Differentiate both sides with respect to (x)

Differentiate the left - hand side: By the chain rule, if (y = \ln u) and (u = f(x)), then (\frac{d}{dx}(\ln f(x))=\frac{f^{\prime}(x)}{f(x)}). Differentiate the right - hand side: Let (u=\ln 4x). First, find (\frac{du}{dx}) using the chain rule. If (u = \ln v) and (v = 4x), then (\frac{du}{dx}=\frac{1}{v}\cdot4=\frac{1}{x}). Now, using the chain rule for (y = u^{2}), (\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}). Substituting (u=\ln 4x) and (\frac{du}{dx}=\frac{1}{x}), we get (\frac{d}{dx}[(\ln 4x)^2]=2\ln(4x)\cdot\frac{1}{x}). So, (\frac{f^{\prime}(x)}{f(x)}=\frac{2\ln(4x)}{x}).

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

Since (f(x)=(4x)^{\ln 4x}), then (f^{\prime}(x)=f(x)\cdot\frac{2\ln(4x)}{x}). Substitute (f(x)) back in: (f^{\prime}(x)=(4x)^{\ln 4x}\cdot\frac{2\ln(4x)}{x}).

Answer:

(f^{\prime}(x)=(4x)^{\ln 4x}\cdot\frac{2\ln(4x)}{x})