let (f(x)=x^{2x}). use logarithmic differentiation to determine the derivative.(f(x)=)(f(1)=)

Explanation:

Step1: Take natural - log of both sides

Let $y = x^{2x}$. Then $\ln y=\ln(x^{2x}) = 2x\ln x$.

Step2: Differentiate both sides with respect to $x$

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

Step3: Solve for $y'$

Multiply both sides by $y$. Since $y = x^{2x}$, we have $y'=x^{2x}(2\ln x + 2)$. So, $f'(x)=x^{2x}(2\ln x + 2)$.

Step4: Evaluate $f'(1)$

Substitute $x = 1$ into $f'(x)$. When $x = 1$, $\ln(1)=0$. Then $f'(1)=1^{2\times1}(2\ln(1)+2)=1\times(0 + 2)=2$.

Answer:

$f'(x)=x^{2x}(2\ln x + 2)$ $f'(1)=2$