Question

attempt 1: 2 attempts remaining. let $f(x)=x^{7x}$. use logarithmic differentiation to determine the derivative. $f(x)=$ $f(1)=$

Explanation:

Step1: Take natural - log of both sides

Take the natural logarithm of \(y = f(x)=x^{7x}\), so \(\ln y=\ln(x^{7x})\). Using the property of logarithms \(\ln(a^{b}) = b\ln(a)\), we get \(\ln y = 7x\ln x\).

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

Differentiate the left - hand side using the chain rule. The derivative of \(\ln y\) with respect to \(x\) is \(\frac{1}{y}y'\). Differentiate the right - hand side using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = 7x\) and \(v=\ln x\). \(u^\prime = 7\) and \(v^\prime=\frac{1}{x}\). So \((7x\ln x)^\prime=7\ln x+7x\cdot\frac{1}{x}=7\ln x + 7\). Then \(\frac{1}{y}y'=7\ln x + 7\).

Step3: Solve for \(y'\)

Multiply both sides by \(y\). Since \(y = x^{7x}\), we have \(y'=f'(x)=x^{7x}(7\ln x + 7)\).

Step4: Find \(f'(1)\)

Substitute \(x = 1\) into \(f'(x)\). Since \(\ln(1)=0\), then \(f'(1)=1^{7\times1}(7\ln(1)+7)=1\times(0 + 7)=7\).

Answer:

\(f'(x)=x^{7x}(7\ln x + 7)\)
\(f'(1)=7\)