use logarithmic differentiation to find the derivative of the func y = x^4x y =

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(x^{4x})$. Using the property $\ln(a^b)=b\ln(a)$, we get $\ln y = 4x\ln x$.

Step2: Differentiate both sides with respect to $x$

The derivative of the left - hand side is $\frac{y'}{y}$ 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, $\frac{y'}{y}=4\ln x + 4$.

Step3: Solve for $y'$

Multiply both sides by $y$. Since $y = x^{4x}$, we have $y'=x^{4x}(4\ln x + 4)$.

Answer:

$x^{4x}(4\ln x + 4)$