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The problem requires us to use the chain rule to find \( \frac{dz}{dt} \), where \( z = xy^9 - x^2y \), with \( x = t^2 + 1 \) and \( y = t^2 - 1 \).

### Steps to solve:

1. **Find partial derivatives of \( z \) with respect to \( x \) and \( y \):**

   \[
   \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(xy^9 - x^2y) = y^9 - 2xy
   \]
   \[
   \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(xy^9 - x^2y) = 9xy^8 - x^2
   \]

2. **Use the chain rule for \( \frac{dz}{dt} \):**
   \[
   \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}
   \]

3. **Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):**

   \[
   \frac{dx}{dt} = \frac{d}{dt}(t^2 + 1) = 2t
   \]
   \[
   \frac{dy}{dt} = \frac{d}{dt}(t^2 - 1) = 2t
   \]

4. **Substitute everything back into the chain rule:**

   \[
   \frac{dz}{dt} = (y^9 - 2xy) \cdot 2t + (9xy^8 - x^2) \cdot 2t
   \]

5. **Substitute the expressions for \( x \) and \( y \) in terms of \( t \):**

   \[
   x = t^2 + 1, \quad y = t^2 - 1
   \]

Thus, the derivative \( \frac{dz}{dt} \) can now be evaluated by plugging in \( x \) and \( y \) and simplifying.

Would you like me to continue with the full simplification?

### Related questions:
1. How do partial derivatives work in multivariable calculus?
2. Can the chain rule be applied to functions with more than two variables?
3. What is the significance of the product rule in differentiation?
4. How can implicit differentiation be used in chain rule problems?
5. How do we differentiate functions raised to a power like \( y^9 \)?

### Tip: 
When using the chain rule, always ensure to carefully account for each variable's dependence on the others to avoid errors.

Use the chain rule to find dz/dt, where z = xy^9 - x^2y, x = t^2 + 1, and y = t^2 - 1.

Learn how to use the chain rule to differentiate the function z = xy^9 - x^2y, with x = t^2 + 1 and y = t^2 - 1. This detailed step-by-step solution covers the chain rule, partial derivatives, and substitution, ideal for advanced high school or college students studying calculus.

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