use logarithmic differentiation to find the derivative of y with respect to t.\ny = (\\sqrt4{t})^t\n\\frac{dy…

Explanation:

Step1: Rewrite the function

First, rewrite $y = (\sqrt[4]{t})^t=t^{\frac{t}{4}}$. Then take the natural - logarithm of both sides: $\ln y=\ln(t^{\frac{t}{4}})=\frac{t}{4}\ln t$.

Step2: Differentiate both sides with respect to $t$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \frac{t}{4}$ and $v=\ln t$. The derivative of $u$ with respect to $t$ is $\frac{1}{4}$, and the derivative of $v$ with respect to $t$ is $\frac{1}{t}$. So, $\frac{1}{y}\frac{dy}{dt}=\frac{1}{4}\ln t+\frac{t}{4}\cdot\frac{1}{t}=\frac{1}{4}\ln t+\frac{1}{4}$.

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

Multiply both sides by $y$ (since $y = t^{\frac{t}{4}}$), we get $\frac{dy}{dt}=y(\frac{1}{4}\ln t+\frac{1}{4})=t^{\frac{t}{4}}(\frac{1}{4}\ln t+\frac{1}{4})$.

Answer:

$t^{\frac{t}{4}}(\frac{1}{4}\ln t+\frac{1}{4})$