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!