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We are given the equation:

\[
w = \frac{x}{y} + \frac{y}{z}
\]
where
\[
x = e^{2t}, \quad y = 2 + \sin(3t), \quad z = 2 + \cos(5t).
\]

### 1. Using the chain rule to find \( \frac{dw}{dt} \) as a function of \( x, y, z, \) and \( t \):

We apply the chain rule to differentiate \( w \) with respect to \( t \).

\[
w = \frac{x}{y} + \frac{y}{z}
\]

Now, differentiate both terms:

1. For the first term \( \frac{x}{y} \), use the quotient rule:
   \[
   \frac{d}{dt} \left( \frac{x}{y} \right) = \frac{y \frac{dx}{dt} - x \frac{dy}{dt}}{y^2}
   \]
2. For the second term \( \frac{y}{z} \), also use the quotient rule:
   \[
   \frac{d}{dt} \left( \frac{y}{z} \right) = \frac{z \frac{dy}{dt} - y \frac{dz}{dt}}{z^2}
   \]

Thus, the derivative \( \frac{dw}{dt} \) becomes:
\[
\frac{dw}{dt} = \frac{y \frac{dx}{dt} - x \frac{dy}{dt}}{y^2} + \frac{z \frac{dy}{dt} - y \frac{dz}{dt}}{z^2}
\]

### Now, compute the individual derivatives:

- \( \frac{dx}{dt} = \frac{d}{dt} \left( e^{2t} \right) = 2e^{2t} = 2x \)
- \( \frac{dy}{dt} = \frac{d}{dt} \left( 2 + \sin(3t) \right) = 3 \cos(3t) \)
- \( \frac{dz}{dt} = \frac{d}{dt} \left( 2 + \cos(5t) \right) = -5 \sin(5t) \)

Thus, the expression for \( \frac{dw}{dt} \) becomes:
\[
\frac{dw}{dt} = \frac{y(2x) - x(3\cos(3t))}{y^2} + \frac{z(3\cos(3t)) - y(-5\sin(5t))}{z^2}
\]

This is the answer for part 1, as requested.

---

### 2. Now, evaluate \( \frac{dw}{dt} \) when \( t = 0 \):

First, substitute \( t = 0 \) into the expressions for \( x, y, \) and \( z \):

- \( x = e^{2(0)} = e^0 = 1 \)
- \( y = 2 + \sin(3(0)) = 2 + 0 = 2 \)
- \( z = 2 + \cos(5(0)) = 2 + 1 = 3 \)

Now substitute \( t = 0 \) into the derivatives:

- \( \frac{dx}{dt} = 2x = 2(1) = 2 \)
- \( \frac{dy}{dt} = 3 \cos(3(0)) = 3(1) = 3 \)
- \( \frac{dz}{dt} = -5 \sin(5(0)) = 0 \)

Substitute these values into the expression for \( \frac{dw}{dt} \):

\[
\frac{dw}{dt} = \frac{2(2) - 1(3)}{2^2} + \frac{3(3) - 2(0)}{3^2}
\]

Simplifying each term:

\[
\frac{dw}{dt} = \frac{4 - 3}{4} + \frac{9 - 0}{9}
\]
\[
\frac{dw}{dt} = \frac{1}{4} + 1 = \frac{1}{4} + \frac{4}{4} = \frac{5}{4}
\]

Thus, when \( t = 0 \), \( \frac{dw}{dt} = \frac{5}{4} \).

---

Let me know if you need further clarification or details!

### Related Questions:
1. How would you extend this to three variables instead of two?
2. What happens to \( \frac{dw}{dt} \) as \( t \to \infty \)?
3. Can you explore the behavior of \( w \) graphically in terms of \( t \)?
4. How would the result change if \( y \) and \( z \) were constants?
5. What happens if \( t = \frac{\pi}{6} \) instead of 0?

### Tip:
When applying the quotient rule, always make sure to differentiate both the numerator and the denominator correctly to avoid mistakes in multi-variable functions!

Suppose w = x/y + y/z, where x = e^{2t}, y = 2 + sin(3t), and z = 2 + cos(5t). Use the chain rule to find dw/dt as a function of x, y, z, and t. Do not rewrite x, y, and z in terms of t. Then, evaluate dw/dt when t = 0.

Solve a calculus problem using the chain rule and quotient rule. Find the derivative dw/dt for the function w = x/y + y/z, where x = e^{2t}, y = 2 + sin(3t), and z = 2 + cos(5t). Includes step-by-step differentiation and evaluation at t = 0.

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