Find the following derivatives. Express your answer in terms of the independent variables. $$ w_{s} $$ and $$ w_{t} $$, where $$ w=\frac{x-z}{3 y+z}, x=3 s+4 t, y=s t $$, and $$ z=3 s-4 t $$ $$ \frac{\partial w}{\partial x}=\frac{1}{3 y+z} $$ $$ (T y p e $$ an expression using $$ x, y $$, and $$ z $$ as the variables.) $$ \frac{\partial w}{\partial y}=-\frac{3(x-z)}{(3 y+z)^{2}} $$ $$ (T y p e $$ an expression using $$ x, y $$, and $$ z $$ as the variables.) $$ \frac{\partial w}{\partial z}=\square $$ (Type an expression using $$ x, y $$, and $$ z $$ as the variables.)

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We start with  
$$
w=\frac{x-z}{3y+z},
$$  
and we want  
$$
\frac{\partial w}{\partial z}
$$  
in terms of $$x$$, $$y$$, and $$z$$. For fixed $$x$$ and $$y$$, we differentiate with respect to $$z$$ using the quotient rule.

Let  
$$
u = x - z \quad\text{and}\quad v = 3y + z.
$$

Then their derivatives with respect to $$z$$ are  
$$
u' = \frac{d}{dz}(x-z) = -1,
$$  
$$
v' = \frac{d}{dz}(3y+z) = 1.
$$

The quotient rule states that  
$$
\frac{\partial w}{\partial z} = \frac{v\,u' - u\,v'}{v^2}.
$$

Substitute in the expressions for $$u$$, $$u'$$, $$v$$, and $$v'$$:  
$$
\frac{\partial w}{\partial z} = \frac{(3y+z)(-1) - (x-z)(1)}{(3y+z)^2}.
$$

Simplify the numerator:  
$$
(3y+z)(-1) - (x-z) = -3y - z - x + z = -x - 3y.
$$

Thus, the derivative is  
$$
\frac{\partial w}{\partial z} = -\frac{x + 3y}{(3y+z)^2}.
$$
$$ \frac{\partial w}{\partial z} = -\frac{x + 3y}{(3y + z)^2} $$

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