Question

derivatives of logarithmic functions: (1 point)
let $f(x)=x^{2x}$.
use logarithmic differentiation to determine the derivative.
$f(x)=square$

Explanation:

Step1: Take natural - log of both sides

Take the natural logarithm of $y = f(x)=x^{2x}$. So, $\ln y=\ln(x^{2x})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln y = 2x\ln x$.

Step2: Differentiate both sides with respect to $x$

Differentiate $\ln y$ with respect to $x$ using the chain - rule. The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}\cdot y'$. The derivative of $2x\ln x$ with respect to $x$ using the product - rule $(uv)' = u'v+uv'$ where $u = 2x$ and $v=\ln x$. So, $u'=2$ and $v'=\frac{1}{x}$, then $(2x\ln x)'=2\ln x + 2x\cdot\frac{1}{x}=2\ln x + 2$. So, $\frac{1}{y}\cdot y'=2\ln x + 2$.

Step3: Solve for $y'$

Multiply both sides by $y$. Since $y = x^{2x}$, we have $y'=y(2\ln x + 2)$. Substituting $y = x^{2x}$ back in, we get $y'=x^{2x}(2\ln x + 2)$.

Answer:

$x^{2x}(2\ln x + 2)$