Question

let ( f(x)=x^{4x} ). use logarithmic differentiation to determine the derivative.
( f^{prime}(x)= )
( f^{prime}(1)= )
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Explanation:

Step1: Take natural logarithm on both sides

Given \(y = x^{4x}\), take \(\ln\) on both sides: \(\ln y=\ln(x^{4x})\).
By the power - rule of logarithms \(\ln a^{b}=b\ln a\), we get \(\ln y = 4x\ln x\).

Step2: Differentiate both sides with respect to \(x\)

Differentiate \(\ln y\) with respect to \(x\) using the chain - rule. \(\frac{d}{dx}(\ln y)=\frac{1}{y}\cdot y'\).
Differentiate \(4x\ln x\) with respect to \(x\) using the product - rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 4x\) and \(v=\ln x\). \(u^\prime=4\) and \(v^\prime=\frac{1}{x}\). So \(\frac{d}{dx}(4x\ln x)=4\ln x + 4x\cdot\frac{1}{x}=4\ln x + 4\).
Then \(\frac{y'}{y}=4\ln x + 4\).

Step3: Solve for \(y'\)

Since \(y = x^{4x}\), multiply both sides of \(\frac{y'}{y}=4\ln x + 4\) by \(y\). We get \(y'=x^{4x}(4\ln x + 4)\).

Step4: Find \(y^\prime(1)\)

Substitute \(x = 1\) into \(y'=x^{4x}(4\ln x + 4)\).
When \(x = 1\), \(x^{4x}=1^{4\times1}=1\) and \(\ln(1)=0\). So \(y^\prime(1)=1\times(4\times0 + 4)=4\).

Answer:

\(f^{\prime}(x)=x^{4x}(4\ln x + 4)\)
\(f^{\prime}(1)=4\)