Question

use logarithmic differentiation to find the derivative of y with respect to t. y = (\sqrt7{t})^t. \frac{dy}{dt}=\square

Explanation:

Step1: Rewrite the function

First, rewrite $y = (\sqrt[7]{t})^t$ as $y=t^{\frac{t}{7}}$. Then take the natural - logarithm of both sides: $\ln y=\ln(t^{\frac{t}{7}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln y = \frac{t}{7}\ln t$.

Step2: Differentiate both sides with respect to $t$

Differentiate the left - hand side using the chain rule. The derivative of $\ln y$ with respect to $t$ is $\frac{1}{y}\cdot\frac{dy}{dt}$. For the right - hand side, use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \frac{t}{7}$ and $v=\ln t$. The derivative of $u=\frac{t}{7}$ with respect to $t$ is $\frac{1}{7}$, and the derivative of $v = \ln t$ with respect to $t$ is $\frac{1}{t}$. So, $(\frac{t}{7}\ln t)^\prime=\frac{1}{7}\ln t+\frac{t}{7}\cdot\frac{1}{t}=\frac{1}{7}\ln t+\frac{1}{7}$.

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

We have $\frac{1}{y}\cdot\frac{dy}{dt}=\frac{1}{7}\ln t+\frac{1}{7}$. Multiply both sides by $y$ to solve for $\frac{dy}{dt}$. Since $y = t^{\frac{t}{7}}$, then $\frac{dy}{dt}=t^{\frac{t}{7}}(\frac{1}{7}\ln t+\frac{1}{7})=\frac{1}{7}t^{\frac{t}{7}}(1 + \ln t)$.

Answer:

$\frac{1}{7}t^{\frac{t}{7}}(1+\ln t)$