Question

Partial derivatives and chain rule problem: Show that\[z _{x}x _{y}y _{z}=-1\] given F(x,y,z)=0 and z=f(x,y) y=g(x,z) x=h(y,z) and
\[F _{x} \neq 0
F _{y} \neq 0
F _{z} \neq 0\]

Answer 1

Show that\[z _{x}x _{y}y _{z}=-1\] given F(x,y,z)=0 and z=f(x,y) y=g(x,z) x=h(y,z) and
\[F _{x} \neq 0
F _{y} \neq 0
F _{z} \neq 0\]

Answer 2

So why aren't you using chain rule?

Answer 3

I believe I'm supposed to expand on the fact that dy/dx=-Fx/Fy

Id like to simply say that Zx=-Fx/Fz (and so on) but I don't think that follows, and I'm not sure how to show that it doesn't

Answer 4

\[
\frac{d}{dx}F(x,y,z) = F_x+F_yy_x+F_zz_x\\
\frac{d}{dy}F(x,y,z) = F_yx_y+F_y+F_zz_y\\
\frac{d}{dz}F(x,y,z) = F_xx_z+F_yy_z+F_z
\]

Answer 5

All of these equations should equal each \(0\) as well.

Answer 6

So you'd say \[
z_x=-\frac{F_x+F_yy_x}{F_z}
\]

Answer 7

I got those equations for Xy Yz and Zx and multiplied them out to see if I could get -1 didn't

Answer 8

What did you get?

Answer 9

FyYzXy+1+FyYx+FxXz+ZyYzXz+FzZy+FzYxZy+FxXzZy

Answer 10

Can you factor it a bit?

Answer 11

FyYz(Xz+1)+FxXz(Zy+1)+FzZy(Yz+1)+ZyYzXz+1

Answer 12

I wonder if \[
F_yy_z=-F_xx_z-F_z
\]Wouldn't help cancel stuff out.

Answer 13

Wait!

Answer 14

You could expand out \(z_yy_zx_z\)

Answer 15

Also, why isn't there a negative sign? Where did it go?

Answer 16

\(\color{blue}{\text{Originally Posted by}}\) @wio
So you'd say \[
z_x=-\frac{F_x+F_yy_x}{F_z}
\]
\(\color{blue}{\text{End of Quote}}\)
Let's start from here.

Answer 17

it is negative

Answer 18

The whole thing is i believe

Answer 19

I think you're on to something swapping things out with the original equations