Question

let (f(x)=x^{5x}). use logarithmic differentiation to determine the derivative (f(x)=square)

Explanation:

Step1: Take natural - log of both sides

Let \(y = f(x)=x^{5x}\). Then \(\ln y=\ln(x^{5x})\). Using the property of logarithms \(\ln(a^b)=b\ln(a)\), we have \(\ln y = 5x\ln x\).

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

The derivative of the left - hand side is \(\frac{1}{y}y'\) by the chain rule. The derivative of the right - hand side using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 5x\) and \(v=\ln x\). \(u^\prime=5\) and \(v^\prime=\frac{1}{x}\). So \((5x\ln x)^\prime=5\ln x + 5x\cdot\frac{1}{x}=5\ln x + 5\). Thus, \(\frac{1}{y}y'=5\ln x + 5\).

Step3: Solve for \(y'\)

Multiply both sides by \(y\). Since \(y = x^{5x}\), we get \(y'=x^{5x}(5\ln x + 5)\).

Answer:

\(x^{5x}(5\ln x + 5)\)