use logarithmic differentiation to find the derivative of y with respect to t.\ny = (\\sqrt{t})^t\n\\frac{dy}…

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(\sqrt{t})^t=t\ln\sqrt{t}=\frac{t}{2}\ln t$

Step2: Differentiate both sides with respect to $t$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = \frac{t}{2}$ and $v=\ln t$. The derivative of $\frac{t}{2}$ with respect to $t$ is $\frac{1}{2}$, and the derivative of $\ln t$ with respect to $t$ is $\frac{1}{t}$. So, $\frac{1}{y}\frac{dy}{dt}=\frac{1}{2}\ln t+\frac{t}{2}\cdot\frac{1}{t}=\frac{1}{2}\ln t+\frac{1}{2}$

Step3: Solve for $\frac{dy}{dt}$

Multiply both sides by $y = (\sqrt{t})^t$ $\frac{dy}{dt}=(\sqrt{t})^t\left(\frac{1}{2}\ln t+\frac{1}{2}\right)$

Answer:

$(\sqrt{t})^t\left(\frac{1}{2}\ln t+\frac{1}{2}\right)$