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

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UpStudy AI · Accepted Answer

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

Using the chain rule, we have
$$
z_s=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s},
$$
and
$$
z_t=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.
$$

The given derivatives are:
$$
\frac{\partial z}{\partial x}=7y-4xy,\quad \frac{\partial z}{\partial y}=7x-2x^2,
$$
$$
\frac{\partial x}{\partial s}=2,\quad \frac{\partial y}{\partial s}=2.
$$

Since
$$
x=2s+5t,\quad y=2s-5t,
$$
we easily compute
$$
\frac{\partial x}{\partial t}=5,\quad \frac{\partial y}{\partial t}=-5.
$$

Now, substituting these into the chain-rule formulas:

1. For $$z_s$$:
   $$
   z_s=(7y-4xy)(2)+(7x-2x^2)(2).
   $$
   Factoring the “2” gives:
   $$
   z_s=2\Bigl[(7y-4xy)+(7x-2x^2)\Bigr]
   $$
   or, if you prefer, you may also leave it as:
   $$
   z_s=14y-8xy+14x-4x^2.
   $$

2. For $$z_t$$:
   $$
   z_t=(7y-4xy)(5)+(7x-2x^2)(-5).
   $$
   Thus,
   $$
   z_t=5(7y-4xy)-5(7x-2x^2)
   $$
   which simplifies to
   $$
   z_t=35y-20xy-35x+10x^2.
   $$

Finally, the missing derivative is
$$
\frac{\partial x}{\partial t}=5.
$$
$$ z_s = 14y - 8xy + 14x - 4x^2 $$ $$ z_t = 35y - 20xy - 35x + 10x^2 $$ $$ \frac{\partial x}{\partial t} = 5 $$

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