(a) Express ∂z
∂u and ∂z
∂y as functions of u and y, both by using the Chain Rule and by expressing

z directly in terms of u and y before differentiating. 

z = 4e
x
ln y, x = ln(u cos v), y = u sin v (u, v) = 
2,
π
4

We are given the following expressions:

\[
z = 4e^x \ln y
\]
\[
x = \ln(u \cos v)
\]
\[
y = u \sin v
\]

where \((u, v) = (2, \frac{\pi}{4})\).

We need to express \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial y}\) in terms of \(u\) and \(y\) using both the **Chain Rule** and by differentiating \(z\) directly with respect to \(u\) and \(y\).

### Part 1: Using the Chain Rule
First, we can apply the chain rule to compute \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial y}\).

#### 1.1 \(\frac{\partial z}{\partial u}\) using the chain rule:

We know that:
\[
z = 4e^x \ln y
\]

By the chain rule:
\[
\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u}
\]

Let's compute the partial derivatives step by step:

- \(\frac{\partial z}{\partial x} = 4e^x \ln y\)
- \(\frac{\partial x}{\partial u} = \frac{1}{u \cos v}\)
- \(\frac{\partial z}{\partial y} = \frac{4e^x}{y}\)
- \(\frac{\partial y}{\partial u} = \sin v\)

Now we substitute all these into the chain rule formula:
\[
\frac{\partial z}{\partial u} = 4e^x \ln y \cdot \frac{1}{u \cos v} + \frac{4e^x}{y} \cdot \sin v
\]

#### 1.2 \(\frac{\partial z}{\partial y}\) using the chain rule:

From \(z = 4e^x \ln y\), we can directly differentiate with respect to \(y\):
\[
\frac{\partial z}{\partial y} = \frac{4e^x}{y}
\]

### Part 2: Expressing \(z\) directly in terms of \(u\) and \(y\)

#### 2.1 Substituting \(x = \ln(u \cos v)\) and \(y = u \sin v\) into \(z\):

We start by substituting \(x\) and \(y\) into the original equation for \(z\):
\[
z = 4e^{\ln(u \cos v)} \ln(u \sin v)
\]
Since \(e^{\ln(a)} = a\), this simplifies to:
\[
z = 4(u \cos v) \ln(u \sin v)
\]
Now, we can differentiate directly with respect to \(u\) and \(y\).

#### 2.2 \(\frac{\partial z}{\partial u}\):

Differentiating \(z = 4(u \cos v) \ln(u \sin v)\) with respect to \(u\):
\[
\frac{\partial z}{\partial u} = 4 \left[ \cos v \ln(u \sin v) + \frac{\cos v}{u \sin v} \cdot u \right]
\]
Simplifying:
\[
\frac{\partial z}{\partial u} = 4\cos v \left( \ln(u \sin v) + 1 \right)
\]

#### 2.3 \(\frac{\partial z}{\partial y}\):

We already computed \(\frac{\partial z}{\partial y}\) using the chain rule, which gives:
\[
\frac{\partial z}{\partial y} = \frac{4e^x}{y}
\]

Now, let's plug in the values \((u, v) = (2, \frac{\pi}{4})\) to simplify and obtain the numeric results.

### Substituting \(u = 2\) and \(v = \frac{\pi}{4}\):

- \(\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)

- \[
x = \ln\left(2 \cdot \frac{1}{\sqrt{2}}\right) = \ln\left(\sqrt{2}\right) = \frac{1}{2} \ln(2)
\]

- \[
y = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}
\]

Now you can substitute these values into the expressions for \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial y}\) to obtain the final numeric results.

Would you like me to compute these numerically for you?

---

**5 Related Questions:**
1. How does the chain rule apply to multivariable functions in general?
2. What is the significance of implicit differentiation in this context?
3. Can you apply the same method for partial derivatives when there are more variables?
4. What are some practical applications of these types of multivariable derivatives?
5. How would the result change if \(z\) had more complicated dependencies on \(x\) and \(y\)?

**Tip:** The chain rule in multivariable calculus helps link how changes in one set of variables affect another through intermediate variables.

Express ∂z/∂u and ∂z/∂y as functions of u and y, both by using the Chain Rule and by expressing z directly in terms of u and y before differentiating. z = 4e^x ln y, x = ln(u cos v), y = u sin v where (u, v) = (2, π/4).

Learn how to compute the partial derivatives ∂z/∂u and ∂z/∂y using the chain rule and direct differentiation. This problem involves multivariable calculus concepts such as the chain rule, implicit differentiation, and partial derivatives, with detailed step-by-step solutions.

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