$$ 1 \leftarrow \quad $$ Find the following derivatives. $$ z_{s} $$ and $$ z_{4} $$, where $$ z=7 x y-2 x^{2} y, x=2 s+5 t $$, and $$ y=2 s-5 t $$ $$ \frac{\partial z}{\partial x}=\square $$ (Type an expression using $$ x $$ and $$ y $$ as the variables.)

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We start with
$$
z = 7xy - 2x^2 y,
$$
with
$$
x = 2s + 5t \quad \text{and} \quad y = 2s - 5t.
$$

Because the question asks for several derivatives, we first compute the partial derivatives of $$ z $$ with respect to $$ x $$ and $$ y $$ (treating the other as constant).

1. Differentiate $$ z $$ with respect to $$ x $$:
   $$
   z_x = \frac{\partial}{\partial x}(7xy) - \frac{\partial}{\partial x}(2x^2 y) = 7y - 4xy.
   $$

2. Differentiate $$ z $$ with respect to $$ y $$:
   $$
   z_y = \frac{\partial}{\partial y}(7xy - 2x^2 y) = 7x - 2x^2.
   $$

Since $$ x $$ and $$ y $$ depend on $$ s $$ and $$ t $$, we can use the chain rule.

For a variable $$ s $$ we have:
$$
z_s = z_x\,x_s + z_y\,y_s.
$$
Since
$$
x = 2s+5t \quad \Longrightarrow \quad x_s=2,
$$
$$
y = 2s-5t \quad \Longrightarrow \quad y_s=2,
$$
it follows that
$$
z_s = (7y-4xy)(2) + (7x-2x^2)(2) = 2(7y-4xy+7x-2x^2).
$$

Similarly, for $$ t $$ (noting that the question writes $$ z_{4} $$ which is most likely a typo for $$ z_t $$):
$$
z_t = z_x\,x_t + z_y\,y_t.
$$
Since
$$
x_t=5 \quad \text{and} \quad y_t=-5,
$$
we obtain
$$
z_t = (7y-4xy)(5) + (7x-2x^2)(-5) = 5(7y-4xy-7x+2x^2).
$$

The last part of the problem asks us to express
$$
\frac{\partial z}{\partial x}
$$
using $$ x $$ and $$ y $$. From our calculation we have:
$$
\frac{\partial z}{\partial x} = 7y-4xy.
$$
$$ \frac{\partial z}{\partial x} = 7y - 4xy $$

\( 1 \leftarrow \quad \) Find the following derivatives. \( z_{s} \) and \( z_{4} \), where \( z=7 x y-2 x^{2} y, x=2 s+5 t \), and \( y=2 s-5 t \) \( \frac{\partial z}{\partial x}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

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