Find $\frac{\partial z}{\partial x}$ if $xy+yz+zx = 1$

Question

Find $\frac{\partial z}{\partial x}$ if $xy+yz+zx = 1$

I don't understand this question at first. It looks like $x,y,z $ are dependent. So we proceed differentiation partially wrt x:

$$xy_x +y + y_x z + z_x y + z_x x+z=0$$

This gives $$z_x = \frac{-z-y-y_xz-xy_x}{x+y}$$

But given answer is $\frac{-z-y}{x+y}$, meaning that they take $y_x = 0$, saying that $y$ and $x$ independent. But how this makes sense, then why not take $z$ and $x$ also dependent and say $z_x = 0$ ?

Please tell me the reasoning! I think that they mean to say treat $y$ as constant when finding $z_x$

Answer 1

You should understand the definition of the partial derivative.

$\frac{\partial z}{\partial x}$ implies "the partial derivative of $z$ with respect to $x$, with other variables held constant.

Hence (holding $y$ as constant, implying its derivative is zero): $$xy_x +y + y_x z + z_x y + z_x x+z=0 \iff \\ x\cdot 0+y+0\cdot z+z_xy+z_xx+z=0 \iff \\ z_x=\frac{-y-z}{x+y}.$$

Answer 2

Variables $x$ and $y$ are independent, but $z$ depends on both $x$ and $y$.

More formally, you can use implicit function theorem. With this theorem, you will get

$$\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

where $F=xy+yz+zx-1$. Of course, given formula is true for points where $\frac{\partial F}{\partial y}\ne 0$.

Answer 3

Hint: You can write $$z=\frac{1-xy}{x+y}$$ and you do not need the implicit function theorem.

Answer 4

As @Dr. Sonnhard Graubner said:

You should take:

$$z=\frac{1-xy}{x+y}$$

Partially differentiating with respect to $x$. we have:

$$\frac{\partial z}{\partial x}=z_x=\frac{(x+y)\frac{\partial}{\partial x}(1-xy)-(1-xy)\frac{\partial}{\partial x}(x+y)}{(x+y)^2}$$

$$z_x=\frac{-(x+y)y-(1-xy)}{(x+y)^2}$$

$$z_x=\frac{-y-\frac{(1-xy)}{x+y}}{x+y}$$

$$z_x=\frac{-y- z}{x+y}$$

Answer 5

We have that

$$xy+yz+zx = 1\implies (y+z)dx+(x+z)dy+(y+x)dz=0$$

then

$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial x}dy=-\frac{y+z}{y+x}dx-\frac{x+z}{y+x}dy$$

and therefore

$$\frac{\partial z}{\partial x}=-\frac{y+z}{y+x}, \quad \frac{\partial z}{\partial y}=-\frac{x+z}{y+x}$$