Question

use logarithmic differentiation to evaluate ( f^{prime}(x) ).

( f(x)=(4 x)^{ln 4 x} )

( f^{prime}(x)= )
(use parentheses to clearly denote the argument of each function.)

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:

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