Partial derivative

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  • #1
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Homework Statement


If w = w(x, y, z) is given implicitly by F(x, y, z, w) = 0, find a formula
for both ∂w/∂z and ∂^2w/∂y∂z . You may assume that each function is sufficiently
differentiable and anything you divide by during the process of your
solution is non-zero.

The Attempt at a Solution


I know in class, my professor showed us how to implicitly differentiate using this method where if y = y(x) and F(x,y) = 0, you can do dF/dx = ∂F/∂x + ∂F/∂y*dy/dx = 0 and dy/dx = -(∂F/∂x)/(∂F/∂y),
Is this the kind of formula my professor wants us to find?
 

Answers and Replies

  • #2
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3
You have to use the chain rule..
 
  • #3
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[itex]0 = \frac{\partial F}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial z} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial z} + \hspace{1 mm}...[/itex]
 
Last edited:
  • #4
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Yea I got ∂w/∂z, but I have no idea how to get the other one, do I do ∂/∂y(∂w/∂z)?
 
  • #5
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Yea I got ∂w/∂z, but I have no idea how to get the other one, do I do ∂/∂y(∂w/∂z)?
Yes
 
  • #6
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I am not sure what ∂/∂y(∂w/∂x*∂x/∂z) would be, would it just be, d^2w/∂x∂y * ∂^2x/∂z∂y?
 
  • #7
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Show us what you got for ##\partial w/\partial z##, and then we can discuss what to do next.

Chet
 
  • #8
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I am not sure what ∂/∂y(∂w/∂x*∂x/∂z) would be, would it just be, d^2w/∂x∂y * ∂^2x/∂z∂y?
I dont think so, I believe you'll need to apply the product rule to the expression that you used to get your first answer..
 
  • #9
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0 = ∂F/∂x + ∂F/∂y + ∂F/∂z + ∂F/∂w*∂w/∂z

∂w/∂z = -(∂F/∂x + ∂F/∂y + ∂F/∂z)/∂F/∂w

Looks like I might needs quotient rule.
 
  • #10
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0 = ∂F/∂x + ∂F/∂y + ∂F/∂z + ∂F/∂w*∂w/∂z

∂w/∂z = -(∂F/∂x + ∂F/∂y + ∂F/∂z)/∂F/∂w

Looks like I might needs quotient rule.
It looks like it needs more than just that. This result is incorrect.

Chet
 
  • #11
RUber
Homework Helper
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Look at post number 3.
Usually it is safe to assume that ##\partial x / \partial z =0##.
 
  • #13
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I don't know what to do at all, I am doing exactly what my prof said to do, make a tree, with F at the top then branch out the different variables then branch out w since it is a function of x,y,z. What exactly did I do wrong in my answer?
 
  • #14
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I don't know what to do at all, I am doing exactly what my prof said to do, make a tree, with F at the top then branch out the different variables then branch out w since it is a function of x,y,z. What exactly did I do wrong in my answer?
I think that you just failed to realize that:
[itex] \frac{\partial F}{\partial x} * 0 = 0 [/itex]...there are some terms in your answer that should not be there...
 
  • #15
andrewkirk
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0 = ∂F/∂x + ∂F/∂y + ∂F/∂z + ∂F/∂w*∂w/∂z

∂w/∂z = -(∂F/∂x + ∂F/∂y + ∂F/∂z)/∂F/∂w
The first three terms of the original differentiation are missing factors. Presumably what you want is
$$
0=\frac{\partial F}{\partial z}=
\frac{\partial F}{\partial x}\frac{\partial x}{\partial z}+
\frac{\partial F}{\partial y}\frac{\partial y}{\partial z}+
\frac{\partial F}{\partial z}\frac{\partial z}{\partial z}
+
\frac{\partial F}{\partial w}\frac{\partial w}{\partial z}$$
Then use the fact that
$$\frac{\partial x}{\partial z}=
\frac{\partial y}{\partial z}=0$$
and
$$\frac{\partial z}{\partial z}=1$$
 
  • #16
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Why are you multiplying them by ∂x/∂z and ∂y/∂z? If you don't mind me asking, I was never taught you had to do that, well maybe I was but I don't remember, I am looking through my notes and it wasn't mentioned. Like I understand with ∂F/∂w, but not the others.
 
  • #17
351
3
Why are you multiplying them by ∂x/∂z and ∂y/∂z? If you don't mind me asking, I was never taught you had to do that, well maybe I was but I don't remember, I am looking through my notes and it wasn't mentioned. Like I understand with ∂F/∂w, but not the others.
It is because you are required to use the chain rule, you are differentiating a composite function of more than one variable..
 
  • #18
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3
Maybe a (somewhat) unrelated example may clear things up:
Suppose you have a function [itex] z = f(x,y) [/itex] where [itex] x = g(s,t) [/itex] and [itex] y=h(s,t) [/itex]. If you want to find the derivative of z with respect to s (a similar situation to the problem you are trying to solve) then you would use the chain rule as follows:
[itex] \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} [/itex]
 
  • #19
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Oh alright I think I understand looking at the definition, but in class all the examples were like where x,y,z where all functions of t, or s and t, and I never realized that if they aren't a function of anything you still have to use chain rule on it.
 
  • #20
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3
You have to use the chain rule because you have:
[itex] w = f(x,y,z) [/itex] given implicitly by [itex] F(x,y,z,w) = 0 [/itex]. So you essentially have [itex] F(x,y,z, f(x,y,z)) = 0 [/itex]. Which is a composite function.
 
  • #21
andrewkirk
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The difference between curly and straight 'd's is a bit tricky in that one. What I wrote above doesn't get them correct because ##\frac{\partial F}{\partial z}## implies that ##w##, as well as ##x,y##, is kept constant, which prevents us from using the fourth term in the way we want. So I think we need to write it as a total derivative:

$$
0=\frac{dF}{dz}=
\frac{\partial F}{\partial x}\frac{dx}{dz}+
\frac{\partial F}{\partial y}\frac{dy}{dz}+
\frac{\partial F}{\partial z}\frac{dz}{dz}
+
\frac{\partial F}{\partial w}\frac{dw}{dz}$$

This then gives us an answer for ##\frac{dw}{dz}## rather than ##\frac{\partial w}{\partial z}##.

But that's OK because

$$\frac{dw}{dz}=
\frac{\partial w}{\partial x}\frac{dx}{dz}+
\frac{\partial w}{\partial y}\frac{dy}{dz}+
\frac{\partial w}{\partial z}\frac{dz}{dz}=0+0+1\times\frac{\partial w}{\partial z}$$
 
  • #22
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Let's take a step back. Are you familiar with the following?
[tex]dF=\left(\frac{\partial F}{\partial x}\right)_{y,z,w}dx+\left(\frac{\partial F}{\partial y}\right)_{x,z,w}dy+\left(\frac{\partial F}{\partial z}\right)_{x,y,w}dz+\left(\frac{\partial F}{\partial w}\right)_{x,y,z}dw[/tex]
But, since F is zero, dF = 0, so
[tex]\left(\frac{\partial F}{\partial x}\right)dx+\left(\frac{\partial F}{\partial y}\right)dy+\left(\frac{\partial F}{\partial z}\right)dz+\left(\frac{\partial F}{\partial w}\right)dw=0[/tex]
So,
[tex]dw=-\frac{\partial F/\partial x}{\partial F/\partial w}dx-\frac{\partial F/\partial y}{\partial F/\partial w}dy-\frac{\partial F/\partial z}{\partial F/\partial w}dz[/tex]

Compare this term for term with
[tex]dw=\left(\frac{\partial w}{\partial x}\right)dx+\left(\frac{\partial w}{\partial y}\right)dy+\left(\frac{\partial w}{\partial z}\right)dz[/tex]

Chet
 

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