The problem in the image asks you to compute $\frac{dz}{dt}$, where $z = f(x, y) = xye^{xy}$, with $x(t) = t$ and $y(t) = \frac{1}{t}$.

Step-by-Step Solution:

We are tasked with finding $\frac{dz}{dt}$, and to do this, we will use the chain rule since $z = f(x, y)$ is a function of $x$ and $y$, which in turn are functions of $t$.

1. Restate the function: $z = xye^{xy}$ where $x = t$ and $y = \frac{1}{t}$.

2. Apply the chain rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$

3. Compute partial derivatives:

   • For $\frac{\partial z}{\partial x}$, treat $y$ as a constant: $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( xye^{xy} \right) = ye^{xy} + xy^2e^{xy}$

   • For $\frac{\partial z}{\partial y}$, treat $x$ as a constant: $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( xye^{xy} \right) = xe^{xy} + x^2ye^{xy}$

4. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

   • Since $x(t) = t$, we have $\frac{dx}{dt} = 1$.
   • Since $y(t) = \frac{1}{t}$, we have: $\frac{dy}{dt} = -\frac{1}{t^2}$

5. Substitute $x = t$ and $y = \frac{1}{t}$ into the partial derivatives:

   • For $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = \frac{1}{t} e^1 + t \left( \frac{1}{t^2} \right) e^1 = \frac{1}{t} e + \frac{e}{t}$ This simplifies to: $\frac{\partial z}{\partial x} = \frac{2e}{t}$

   • For $\frac{\partial z}{\partial y}$: $\frac{\partial z}{\partial y} = t e^1 + t^2 \left( \frac{1}{t} \right) e^1 = te + te$ This simplifies to: $\frac{\partial z}{\partial y} = 2te$

6. Substitute into the chain rule expression: $\frac{dz}{dt} = \left( \frac{2e}{t} \right)(1) + (2te) \left( -\frac{1}{t^2} \right)$ Simplify: $\frac{dz}{dt} = \frac{2e}{t} - \frac{2e}{t}$ Thus: $\frac{dz}{dt} = 0$

Final Answer:

$\frac{dz}{dt} = 0$

Would you like more details or further explanations?

Related Questions:

1. How does the chain rule apply to functions of multiple variables?
2. What is the significance of partial derivatives in finding the total derivative?
3. Can you explain why $\frac{dz}{dt} = 0$ in this problem?
4. How would the solution change if $y(t)$ was a different function of $t$?
5. How do exponential functions affect differentiation?

Tip:

Always keep track of how each variable depends on the independent variable when using the chain rule with functions of multiple variables!