Vtyjkyuk

Find $fracpartial zpartial x$ 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 $$z_x = frac-z-y-y_xz-xy_xx+y$$

But given answer is $frac-z-yx+y$, 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 $z_x$

You should understand the definition of the partial derivative.

$fracpartial zpartial x$ 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):
$$xy_x +y + y_x z + z_x y + z_x x+z=0 iff \
xcdot 0+y+0cdot z+z_xy+z_xx+z=0 iff \
z_x=frac-y-zx+y.$$

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

$$fracpartial zpartial x = -fracfracpartial Fpartial xfracpartial Fpartial y$$

where $F=xy+yz+zx-1$. Of course, given formula is true for points where $fracpartial Fpartial yne 0$.

Hint: You can write $$z=frac1-xyx+y$$ and you do not need the implicit function theorem.

As @Dr. Sonnhard Graubner said:

You should take:

$$z=frac1-xyx+y$$

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

$$fracpartial zpartial x=z_x=frac(x+y)fracpartialpartial x(1-xy)-(1-xy)fracpartialpartial x(x+y)(x+y)^2$$

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

$$z_x=frac-y-frac(1-xy)x+yx+y$$

$$z_x=frac-y- zx+y$$

We have that

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

then

$$dz=fracpartial zpartial xdx+fracpartial zpartial xdy=-fracy+zy+xdx-fracx+zy+xdy$$

and therefore

$$fracpartial zpartial x=-fracy+zy+x, quad fracpartial zpartial y=-fracx+zy+x$$