z=(x+3y)ex+y,x=u,y=ln(v),
  find ∂z∂u
 and ∂z∂v
. The variables are restricted to domains on which the functions are defined.

   ∂z∂u=
 

   ∂z∂v=

We are given the function \( z = (x + 3y)e^x + y \), where \( x = u \) and \( y = \ln(v) \). We need to find the partial derivatives of \( z \) with respect to \( u \) and \( v \).

### Step 1: Express \( z \) in terms of \( u \) and \( v \)

We substitute \( x = u \) and \( y = \ln(v) \) into the expression for \( z \):

\[
z = (u + 3\ln(v))e^u + \ln(v)
\]

### Step 2: Find \( \frac{\partial z}{\partial u} \)

To differentiate \( z \) with respect to \( u \), treat \( v \) as a constant. So, the derivative of each term is:

\[
\frac{\partial}{\partial u} \left( (u + 3\ln(v))e^u + \ln(v) \right) = \frac{\partial}{\partial u} \left( (u + 3\ln(v))e^u \right) + \frac{\partial}{\partial u} \left( \ln(v) \right)
\]

Since \( \ln(v) \) is independent of \( u \), its derivative is zero. Now, applying the product rule to the first term:

\[
\frac{\partial}{\partial u} \left( (u + 3\ln(v))e^u \right) = \left( 1 \right)e^u + (u + 3\ln(v))e^u
\]

Simplifying:

\[
\frac{\partial z}{\partial u} = e^u + (u + 3\ln(v))e^u = (u + 1 + 3\ln(v))e^u
\]

### Step 3: Find \( \frac{\partial z}{\partial v} \)

To differentiate \( z \) with respect to \( v \), treat \( u \) as a constant and use the chain rule:

\[
\frac{\partial}{\partial v} \left( (u + 3\ln(v))e^u + \ln(v) \right) = 3 \cdot \frac{1}{v}e^u + \frac{1}{v}
\]

Simplifying:

\[
\frac{\partial z}{\partial v} = \frac{3e^u}{v} + \frac{1}{v} = \frac{3e^u + 1}{v}
\]

### Final Answers

\[
\frac{\partial z}{\partial u} = (u + 1 + 3\ln(v))e^u
\]
\[
\frac{\partial z}{\partial v} = \frac{3e^u + 1}{v}
\]

Let me know if you'd like further details or explanations on any part!

---

Here are 5 related questions:
1. How does the product rule apply to functions involving exponentials?
2. What are the domains of \( u \) and \( v \) in the given problem?
3. How does the chain rule work when differentiating logarithmic functions?
4. What if \( y \) were a more complex function of \( v \)? How would that affect \( \frac{\partial z}{\partial v} \)?
5. What are common applications of partial derivatives in multivariable calculus?

**Tip:** When differentiating composite functions like \( \ln(v) \) or \( e^u \), always carefully apply the chain rule to account for nested dependencies!

The query submitted by the user: z=(x+3y)e^x+y, x=u, y=ln(v), find ∂z/∂u and ∂z/∂v. The variables are restricted to domains on which the functions are defined.

Learn how to calculate the partial derivatives ∂z/∂u and ∂z/∂v for the function z = (x + 3y)e^x + y, with substitutions x = u and y = ln(v). The solution involves applying the product rule and chain rule, using logarithmic and exponential functions. Suitable for undergraduate calculus students.

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