The equation provided is:

$y = \left( \sqrt{t} \right)^t = t^{t/2}$

We are tasked with finding the derivative of $y$ with respect to $t$ using logarithmic differentiation.

Steps for Logarithmic Differentiation:

1. Take the natural logarithm of both sides: $\ln(y) = \ln\left(t^{t/2}\right)$

2. Apply the logarithmic property $\ln(a^b) = b \ln(a)$: $\ln(y) = \frac{t}{2} \ln(t)$

3. Differentiate both sides with respect to $t$. Remember that $\frac{d}{dt}[\ln(y)] = \frac{1}{y} \frac{dy}{dt}$ by implicit differentiation: $\frac{1}{y} \frac{dy}{dt} = \frac{d}{dt}\left(\frac{t}{2} \ln(t)\right)$

4. Use the product rule to differentiate the right-hand side. The product rule states $\frac{d}{dt}[u v] = u'v + uv'$, where $u = \frac{t}{2}$ and $v = \ln(t)$: $\frac{dy}{dt} = y \cdot \left[\frac{1}{2} \ln(t) + \frac{t}{2} \cdot \frac{1}{t}\right]$

   Simplifying: $\frac{dy}{dt} = y \left(\frac{1}{2} \ln(t) + \frac{1}{2}\right)$

5. Substitute $y = t^{t/2}$ back into the equation: $\frac{dy}{dt} = t^{t/2} \left(\frac{1}{2} \ln(t) + \frac{1}{2}\right)$

Thus, the derivative of $y$ with respect to $t$ is:

$\frac{dy}{dt} = t^{t/2} \left(\frac{1}{2} \ln(t) + \frac{1}{2}\right)$

Would you like more details on any part of the process?

Here are 5 related questions you could explore:

1. How does logarithmic differentiation simplify the process for functions with exponents?
2. What is the product rule in differentiation, and how is it applied in this problem?
3. How would you differentiate a function like $y = t^t$ without logarithmic differentiation?
4. What happens when $t = 1$ in the function $y = t^{t/2}$?
5. How do you interpret the derivative of exponential functions with variable exponents?

Tip:

When differentiating logarithmic expressions, always use the chain rule carefully, especially if the logarithm involves a function of $t$.