Question

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

Explanation:

Step1: Take natural - log of both sides

Let $y = f(x)=x^{4x}$. Then $\ln y=\ln(x^{4x})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln y = 4x\ln x$.

Step2: Differentiate both sides with respect to $x$

The derivative of the left - hand side is $\frac{1}{y}y'$ by the chain rule. For the right - hand side, use the product rule $(uv)' = u'v+uv'$, where $u = 4x$ and $v=\ln x$. $u' = 4$ and $v'=\frac{1}{x}$. So $(4x\ln x)'=4\ln x + 4x\cdot\frac{1}{x}=4\ln x + 4$. Thus, $\frac{1}{y}y'=4\ln x + 4$.

Step3: Solve for $y'$

Multiply both sides by $y$ to get $y'=y(4\ln x + 4)$. Since $y = x^{4x}$, then $f'(x)=x^{4x}(4\ln x + 4)$.

Step4: Find $f'(1)$

Substitute $x = 1$ into $f'(x)$. Since $\ln(1)=0$, then $f'(1)=1^{4\times1}(4\ln(1)+4)=1\times(0 + 4)=4$.

Answer:

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