Assignment: find $$ \frac{d z}{d x} $$ and $$ \frac{d z}{d y} $$ for the following: $$ 1-z=(4 x-2 y)(3 x+5 y) $$ $$ 2-\frac{5 x+y}{x-2 y} $$

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**Problem 1**

We are given
$$
1-z = (4x-2y)(3x+5y).
$$
This can be rewritten as
$$
z = 1 - (4x-2y)(3x+5y).
$$

*Step 1. Compute $$\frac{\partial z}{\partial x}$$.*

Differentiate term‐by‐term with respect to $$x$$ (treating $$y$$ as a constant):

The constant $$1$$ gives $$0$$. For the product, use the product rule. Define
$$
u(x,y)=4x-2y\quad\text{and}\quad v(x,y)=3x+5y.
$$
Then,
$$
\frac{\partial u}{\partial x}=4,\quad \frac{\partial v}{\partial x}=3.
$$
By the product rule,
$$
\frac{\partial}{\partial x} \left[u\,v\right] =
\frac{\partial u}{\partial x}\,v + u\,\frac{\partial v}{\partial x}
= 4(3x+5y) + (4x-2y)(3).
$$
Simplify:
$$
4(3x+5y)=12x+20y,\quad (4x-2y)(3)=12x-6y,
$$
so,
$$
\frac{\partial}{\partial x} \left[u\,v\right]
= (12x+20y) + (12x-6y)
= 24x+14y.
$$
Since
$$
z = 1 - (u\,v),
$$
we have
$$
\frac{\partial z}{\partial x} = -\left(24x+14y\right).
$$

*Step 2. Compute $$\frac{\partial z}{\partial y}$$.*

Differentiate with respect to $$y$$ (treating $$x$$ as a constant):

Now,
$$
\frac{\partial u}{\partial y}=-2,\quad \frac{\partial v}{\partial y}=5.
$$
Then, the derivative of the product is
$$
\frac{\partial}{\partial y} \left[u\,v\right]
= \frac{\partial u}{\partial y}\,v + u\,\frac{\partial v}{\partial y}
= (-2)(3x+5y) + (4x-2y)(5).
$$
Simplify:
$$
(-2)(3x+5y)=-6x-10y,\quad (4x-2y)(5)=20x-10y,
$$
so,
$$
\frac{\partial}{\partial y} \left[u\,v\right]
= (-6x-10y) + (20x-10y)
= 14x-20y.
$$
Thus,
$$
\frac{\partial z}{\partial y} = -\left(14x-20y\right) = -14x+20y.
$$

---

**Problem 2**

We are given
$$
z = 2-\frac{5x+y}{x-2y}.
$$

Let
$$
w(x,y)=\frac{5x+y}{x-2y}.
$$
Then,
$$
z=2-w(x,y).
$$
Differentiate using the quotient rule.

Define:
$$
N(x,y)=5x+y,\quad D(x,y)=x-2y.
$$
Then,
$$
\frac{\partial N}{\partial x}=5,\quad \frac{\partial D}{\partial x}=1,
$$
$$
\frac{\partial N}{\partial y}=1,\quad \frac{\partial D}{\partial y}=-2.
$$

*Step 1. Compute $$\frac{\partial z}{\partial x}$$.*

First, find the partial derivative of $$w$$ with respect to $$x$$:
$$
\frac{\partial w}{\partial x} = \frac{D\frac{\partial N}{\partial x}-N\frac{\partial D}{\partial x}}{D^2}
=\frac{(x-2y)(5)- (5x+y)(1)}{(x-2y)^2}.
$$
Simplify the numerator:
$$
5(x-2y)=5x-10y,\quad\text{and}\quad 5x+y,
$$
so,
$$
5x-10y-(5x+y)=-11y.
$$
Thus,
$$
\frac{\partial w}{\partial x}=\frac{-11y}{(x-2y)^2}.
$$
Since $$z=2-w$$, we have
$$
\frac{\partial z}{\partial x}=-\frac{\partial w}{\partial x} = -\left(\frac{-11y}{(x-2y)^2}\right)
=\frac{11y}{(x-2y)^2}.
$$

*Step 2. Compute $$\frac{\partial z}{\partial y}$$.*

Now differentiate $$w$$ with respect to $$y$$:
$$
\frac{\partial w}{\partial y} = \frac{D\frac{\partial N}{\partial y}-N\frac{\partial D}{\partial y}}{D^2}
=\frac{(x-2y)(1)- (5x+y)(-2)}{(x-2y)^2}.
$$
Simplify the numerator:
$$
(x-2y)(1)=x-2y,\quad (5x+y)(-2)=-10x-2y,
$$
so,
$$
x-2y - (-10x-2y) = x-2y+10x+2y =11x.
$$
Thus,
$$
\frac{\partial w}{\partial y}=\frac{11x}{(x-2y)^2}.
$$
Then,
$$
\frac{\partial z}{\partial y}=-\frac{\partial w}{\partial y} = -\frac{11x}{(x-2y)^2}.
$$

---

**Final Answers**

For
$$
z=1-(4x-2y)(3x+5y):
$$
$$
\frac{\partial z}{\partial x} = -\left(24x+14y\right),\quad \frac{\partial z}{\partial y} = -14x+20y.
$$

For
$$
z=2-\frac{5x+y}{x-2y}:
$$
$$
\frac{\partial z}{\partial x} = \frac{11y}{(x-2y)^2},\quad \frac{\partial z}{\partial y} = -\frac{11x}{(x-2y)^2}.
$$
For the first equation: $$ \frac{dz}{dx} = -24x - 14y,\quad \frac{dz}{dy} = -14x + 20y $$ For the second equation: $$ \frac{dz}{dx} = \frac{11y}{(x - 2y)^2},\quad \frac{dz}{dy} = -\frac{11x}{(x - 2y)^2} $$

Assignment: find \( \frac{d z}{d x} \) and \( \frac{d z}{d y} \) for the following: \( 1-z=(4 x-2 y)(3 x+5 y) \) \( 2-\frac{5 x+y}{x-2 y} \)

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