Use differentials to approximate the change in z for the given change in the independent variables. 
zequalsln left parenthesis x Superscript 8 Baseline y right parenthesis when left parenthesis x comma y right parenthesis changes from ​(negative 2​,2​) to ​(negative 1.97​,1.96​)
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Part 1
dzequals
  
enter your response here ​(Type an integer or a​ decimal.)

We are tasked with using differentials to approximate the change in \( z \), where \( z = \ln(x^8 y) \), as the independent variables \( x \) and \( y \) change from \( (x, y) = (-2, 2) \) to \( (x, y) = (-1.97, 1.96) \).

### Step 1: Compute partial derivatives of \( z \)

Given \( z = \ln(x^8 y) \), we need to compute the partial derivatives of \( z \) with respect to \( x \) and \( y \).

#### Partial derivative with respect to \( x \):
\[
z = \ln(x^8 y) = \ln(x^8) + \ln(y) = 8\ln(x) + \ln(y)
\]
Now, the partial derivative of \( z \) with respect to \( x \):
\[
\frac{\partial z}{\partial x} = 8 \cdot \frac{1}{x} = \frac{8}{x}
\]

#### Partial derivative with respect to \( y \):
The partial derivative of \( z \) with respect to \( y \) is:
\[
\frac{\partial z}{\partial y} = \frac{1}{y}
\]

### Step 2: Use differentials to approximate \( dz \)

The total differential of \( z \), \( dz \), is given by:
\[
dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy
\]

Substituting the partial derivatives:
\[
dz = \frac{8}{x} \, dx + \frac{1}{y} \, dy
\]

### Step 3: Evaluate at the initial point \( (x_0, y_0) = (-2, 2) \)

Now, let's evaluate the partial derivatives at the point \( (x_0, y_0) = (-2, 2) \):
\[
\frac{8}{x_0} = \frac{8}{-2} = -4
\]
\[
\frac{1}{y_0} = \frac{1}{2} = 0.5
\]

Thus, the differential becomes:
\[
dz = (-4) \, dx + (0.5) \, dy
\]

### Step 4: Compute \( dx \) and \( dy \)

Now, let's compute the small changes in \( x \) and \( y \):
\[
dx = x_1 - x_0 = -1.97 - (-2) = 0.03
\]
\[
dy = y_1 - y_0 = 1.96 - 2 = -0.04
\]

### Step 5: Approximate \( dz \)

Substitute \( dx = 0.03 \) and \( dy = -0.04 \) into the expression for \( dz \):
\[
dz = (-4)(0.03) + (0.5)(-0.04)
\]
\[
dz = -0.12 - 0.02 = -0.14
\]

Thus, the approximate change in \( z \) is:
\[
dz \approx -0.14
\]

Let me know if you would like more details or have any questions!

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Here are 5 related questions you can try:
1. Use differentials to approximate the change in \( z \) when \( z = x^2 + y^2 \) and \( (x, y) \) changes from \( (1, 1) \) to \( (1.01, 1.02) \).
2. Find the partial derivatives of \( z = \ln(x^3 y^2) \).
3. Approximate the change in \( z = e^{xy} \) when \( (x, y) \) changes from \( (0.5, 1) \) to \( (0.52, 0.98) \).
4. Evaluate the differential \( dz \) when \( z = x^2y + 3x \), \( x = 2 \), and \( y = 3 \).
5. Find the linear approximation of \( f(x, y) = x^2 - y^2 \) near \( (1, 1) \).

**Tip**: When using differentials, you're essentially using a tangent plane to approximate small changes, making it a powerful tool for estimation!

Use differentials to approximate the change in z for the given change in the independent variables. z = ln(x^8 * y) when (x, y) changes from (-2, 2) to (-1.97, 1.96).

Learn how to use differentials to approximate the change in z when z = ln(x^8 * y), and (x, y) changes from (-2, 2) to (-1.97, 1.96). This solution explores the use of partial derivatives and the chain rule for estimating small changes. Ideal for college calculus students.

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