We are tasked with finding the derivative of $u$ with respect to $t$, given the equation $u = XY + yz + zx$, where $X = \frac{1}{t}$, $y = e^t$, and $z = e^{-t}$.

Step-by-step Solution:

1. Rewrite the expression for $u$: $u = XY + yz + zx$ Substitute the given values for $X$, $y$, and $z$: $u = \left(\frac{1}{t}\right)\left(e^t\right) + \left(e^t\right)\left(e^{-t}\right) + \left(e^{-t}\right)\left(\frac{1}{t}\right)$ Simplify the terms: $u = \frac{e^t}{t} + 1 + \frac{e^{-t}}{t}$

2. Differentiate $u$ with respect to $t$:

   Now, apply the product rule and the chain rule where necessary to find $\frac{du}{dt}$.

   $\frac{du}{dt} = \frac{d}{dt} \left(\frac{e^t}{t}\right) + \frac{d}{dt}(1) + \frac{d}{dt} \left(\frac{e^{-t}}{t}\right)$

   Let's differentiate each term separately:

   • For the first term $\frac{e^t}{t}$, apply the quotient rule: $\frac{d}{dt} \left(\frac{e^t}{t}\right) = \frac{t \cdot \frac{d}{dt}(e^t) - e^t \cdot \frac{d}{dt}(t)}{t^2} = \frac{t e^t - e^t}{t^2} = \frac{e^t(t - 1)}{t^2}$

   • The derivative of the constant $1$ is: $\frac{d}{dt}(1) = 0$

   • For the third term $\frac{e^{-t}}{t}$, again apply the quotient rule: $\frac{d}{dt} \left(\frac{e^{-t}}{t}\right) = \frac{t \cdot \frac{d}{dt}(e^{-t}) - e^{-t} \cdot \frac{d}{dt}(t)}{t^2} = \frac{t(-e^{-t}) - e^{-t}}{t^2} = \frac{-e^{-t}(t + 1)}{t^2}$

3. Combine the results:

   Now, sum the derivatives: $\frac{du}{dt} = \frac{e^t(t - 1)}{t^2} + 0 + \frac{-e^{-t}(t + 1)}{t^2}$ Combine the terms into a single fraction: $\frac{du}{dt} = \frac{e^t(t - 1) - e^{-t}(t + 1)}{t^2}$

Thus, the derivative of $u$ with respect to $t$ is: $\frac{du}{dt} = \frac{e^t(t - 1) - e^{-t}(t + 1)}{t^2}$

Would you like any further clarifications or more details on any step?

Here are some related questions:

1. How do you apply the product rule in differentiation?
2. What is the quotient rule, and how does it work?
3. How can you simplify expressions involving exponentials after differentiation?
4. What are the applications of chain rule in calculus?
5. How would the result change if $z$ were a different function of $t$?

Tip: When dealing with derivatives of quotient expressions, it's often useful to simplify terms first before applying the quotient rule.