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Use the Chain Rule to find $$d z / d t$$
$$z=x y^{3}-x^{2} y, \quad x=t^{2}+1, \quad y=t^{2}-1$$

$$z=x y^{3}-x^{2} y, \quad x=t^{2}+1, \quad y=t^{2}-1$$

$$\frac{d z}{d t}=\frac{\partial z}{\partial y} \frac{d y}{d t}+\frac{\partial z}{\partial x} \frac{d x}{d t} $$

$$\frac{d z}{d t}=\left[3 y^{2} x-x^{2}\right] (2 t)+\left[y^{3}-2 yx\right] (2 t)$$

$$\frac{d z}{d t}=2 t\left[3 y^{2} x-x^{2}+y^{3}-2 x y\right]$$

Use the Chain Rule to find $$\partial z / \partial s$$ and $$\partial z / \partial t$$
$$\quad z=(x-y)^{5}, \quad x=s^{2} t, \quad y=s t^{2}$$

(1) $$\frac{d z}{d t}=\frac{\partial z}{\partial y} \frac{\partial y}{\partial t}+\frac{\partial z}{\partial x} \frac{\partial x}{\partial t} $$

$$\frac{\partial z}{\partial t}=5(x-y)^{4}[-1](2 t s)+5(x-y)^{4}\left(s^{2}\right)=5(x-y)^{4} s^{2}-2 s t$$

(2) $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial z}{\partial x} \frac{\partial x}{\partial s} $$

$$\frac{\partial z}{\partial s}=-5(x-y)^{4} t^{2}+5(x-y)^{4}(2 s t)$$

$$\frac{\partial z}{\partial s}=5(x-y)^{4}\left[2 s t-t^{2}\right]$$

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