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 $y = (\sqrt[7]{t})^t=t^{\frac{t}{7}}$. Then take the natural - logarithm of both sides: $\ln y=\ln(t^{\frac{t}{7}})=\frac{t}{7}\ln t$.

Step2: Differentiate both sides with respect to $t$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \frac{t}{7}$ and $v=\ln t$. The derivative of $u$ with respect to $t$ is $\frac{1}{7}$, and the derivative of $v$ with respect to $t$ is $\frac{1}{t}$. So, $\frac{1}{y}\frac{dy}{dt}=\frac{1}{7}\ln t+\frac{t}{7}\cdot\frac{1}{t}=\frac{1}{7}\ln t+\frac{1}{7}$.

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

Multiply both sides by $y$ (since $y = t^{\frac{t}{7}}$), we get $\frac{dy}{dt}=y(\frac{1}{7}\ln t+\frac{1}{7})=t^{\frac{t}{7}}(\frac{1}{7}\ln t+\frac{1}{7})$.

Answer:

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