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

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.