Can you solve this differentiability problem involving implicit functions?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
The discussion revolves around a differentiability problem involving implicit functions defined by the equation F(x,y,z)=0. Participants are tasked with demonstrating that the product of the partial derivatives \(\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-1\) under the condition that F is differentiable and its partial derivatives are nonzero. Several users, including M R, MarkFL, and Opalg, successfully solved the problem, with Opalg providing a detailed solution. The thread emphasizes the importance of understanding implicit differentiation and the relationships between the variables. The problem serves as a practical application of these concepts in multivariable calculus.
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

Problem: Suppose that the equation $F(x,y,z)=0$ implicitly defines each of the three variables $x$, $y$, and $z$ as functions of the other two: $z=f(x,y)$, $y=g(x,z)$ and $x=h(y,z)$. If $F$ is differentiable and $F_x$, $F_y$, and $F_z$ are nonzero, show that
\[\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-1.\]

-----

 
Physics news on Phys.org
This week's problem was correctly answered by M R, MarkFL, and Opalg. You can find Opalg's solution below.

With $z = f(x,y)$, differentiate the equation $F(x,y,f(x,y)) = 0$ partially with respect to $x$, keeping $y$ constant and using the chain rule: $$F_x + \frac{\partial z}{\partial x}F_z = 0$$. Hence $$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$$.

In the same way, $$\frac{\partial x}{\partial y} = -\frac{F_y}{F_x}$$ and $$\frac{\partial y}{\partial z} = -\frac{F_z}{F_y}$$. Therefore $$\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z} = -\frac{F_xF_yF_z}{F_zF_xF_y} = -1$$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K