Find $$ \partial w / \partial s $$ and $$ \partial w / \partial t $$ by using the appropriate Chain Rule. $$ w=x \cos (y z), x=s^{2}, y=t^{2}, z=s-2 t $$

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We are given
$$
w=x\cos(yz),\quad x=s^2,\quad y=t^2,\quad z=s-2t.
$$
We wish to compute $$\frac{\partial w}{\partial s}$$ and $$\frac{\partial w}{\partial t}$$ using the Chain Rule.

**Step 1. Compute the partial derivatives of $$w$$ with respect to $$x$$, $$y$$, and $$z$$.**

Since
$$
w=x\cos(yz),
$$
we have
$$
\frac{\partial w}{\partial x}=\cos(yz),
$$
$$
\frac{\partial w}{\partial y}=\frac{\partial}{\partial y}[x\cos(yz)]
=\,-x\sin(yz)\cdot z,
$$
$$
\frac{\partial w}{\partial z}=\frac{\partial}{\partial z}[x\cos(yz)]
=\,-x\sin(yz)\cdot y.
$$

**Step 2. Compute the derivatives of the intermediate variables $$x$$, $$y$$, and $$z$$ with respect to $$s$$ and $$t$$.**

We have:
$$
x=s^2,\quad y=t^2,\quad z=s-2t.
$$
Thus,
$$
\frac{\partial x}{\partial s}=2s,\quad \frac{\partial x}{\partial t}=0,
$$
$$
\frac{\partial y}{\partial s}=0,\quad \frac{\partial y}{\partial t}=2t,
$$
$$
\frac{\partial z}{\partial s}=1,\quad \frac{\partial z}{\partial t}=-2.
$$

**Step 3. Apply the Chain Rule.**

The chain rule tells us that
$$
\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}.
$$
Substitute the expressions found:
$$
\frac{\partial w}{\partial s}= \cos(yz)(2s) + \Bigl[-x\sin(yz)\cdot z\Bigr](0) + \Bigl[-x\sin(yz)\cdot y\Bigr](1).
$$
Thus,
$$
\frac{\partial w}{\partial s}=2s\cos(yz)-x\,y\sin(yz).
$$
Now replace back $$x$$, $$y$$, and $$z$$ in terms of $$s$$ and $$t$$:
$$
x=s^2,\quad y=t^2,\quad z=s-2t,
$$
so
$$
\frac{\partial w}{\partial s}=2s\cos\bigl(t^2(s-2t)\bigr)-s^2t^2\sin\bigl(t^2(s-2t)\bigr).
$$

Similarly, for $$t$$ we have:
$$
\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}.
$$
Substitute the expressions:
$$
\frac{\partial w}{\partial t}= \cos(yz)(0) + \Bigl[-x\sin(yz)\cdot z\Bigr](2t) + \Bigl[-x\sin(yz)\cdot y\Bigr](-2).
$$
This gives
$$
\frac{\partial w}{\partial t}=-2t\,x\,z\sin(yz)+2x\,y\sin(yz).
$$
Replacing back $$x$$, $$y$$, and $$z$$:
$$
\frac{\partial w}{\partial t}=-2t\,s^2\,(s-2t)\sin\bigl(t^2(s-2t)\bigr) + 2s^2\,t^2\sin\bigl(t^2(s-2t)\bigr).
$$
We may factor the common terms:
$$
\frac{\partial w}{\partial t}=2s^2\sin\bigl(t^2(s-2t)\bigr)\Bigl[t^2-t(s-2t)\Bigr].
$$
Simplify the expression inside the bracket:
$$
t^2-t(s-2t)=t^2-ts+2t^2=3t^2-ts.
$$
Thus,
$$
\frac{\partial w}{\partial t}=2s^2\bigl(3t^2-ts\bigr)\sin\bigl(t^2(s-2t)\bigr).
$$

**Final Answers:**

$$
\boxed{\frac{\partial w}{\partial s}=2s\cos\bigl(t^2(s-2t)\bigr)-s^2t^2\sin\bigl(t^2(s-2t)\bigr)}
$$
$$
\boxed{\frac{\partial w}{\partial t}=-2t\,s^2(s-2t)\sin\bigl(t^2(s-2t)\bigr)+2s^2t^2\sin\bigl(t^2(s-2t)\bigr)}
$$

Any algebraically equivalent form is acceptable.
$$ \frac{\partial w}{\partial s} = 2s \cos(t^2(s - 2t)) - s^2 t^2 \sin(t^2(s - 2t)) $$ $$ \frac{\partial w}{\partial t} = 2s^2 (3t^2 - t s) \sin(t^2(s - 2t)) $$

Find \( \partial w / \partial s \) and \( \partial w / \partial t \) by using the appropriate Chain Rule. \[ w=x \cos (y z), x=s^{2}, y=t^{2}, z=s-2 t \]

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