# Partial derivative

1. Nov 2, 2014

### Panphobia

1. The problem statement, all variables and given/known data
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.

3. 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?

2. Nov 2, 2014

### _N3WTON_

You have to use the chain rule..

3. Nov 2, 2014

### _N3WTON_

$0 = \frac{\partial F}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial z} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial z} + \hspace{1 mm}...$

Last edited: Nov 2, 2014
4. Nov 2, 2014

### Panphobia

Yea I got ∂w/∂z, but I have no idea how to get the other one, do I do ∂/∂y(∂w/∂z)?

5. Nov 2, 2014

Yes

6. Nov 2, 2014

### Panphobia

I am not sure what ∂/∂y(∂w/∂x*∂x/∂z) would be, would it just be, d^2w/∂x∂y * ∂^2x/∂z∂y?

7. Nov 2, 2014

### Staff: Mentor

Show us what you got for $\partial w/\partial z$, and then we can discuss what to do next.

Chet

8. Nov 2, 2014

### _N3WTON_

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. Nov 2, 2014

### Panphobia

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. Nov 3, 2014

### Staff: Mentor

It looks like it needs more than just that. This result is incorrect.

Chet

11. Nov 3, 2014

### RUber

Look at post number 3.
Usually it is safe to assume that $\partial x / \partial z =0$.

12. Nov 3, 2014

### Staff: Mentor

Especially if x is one of the parameters that is being held constant.

13. Nov 3, 2014

### Panphobia

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. Nov 3, 2014

### _N3WTON_

I think that you just failed to realize that:
$\frac{\partial F}{\partial x} * 0 = 0$...there are some terms in your answer that should not be there...

15. Nov 3, 2014

### andrewkirk

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. Nov 3, 2014

### Panphobia

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. Nov 3, 2014

### _N3WTON_

It is because you are required to use the chain rule, you are differentiating a composite function of more than one variable..

18. Nov 3, 2014

### _N3WTON_

Maybe a (somewhat) unrelated example may clear things up:
Suppose you have a function $z = f(x,y)$ where $x = g(s,t)$ and $y=h(s,t)$. 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:
$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$

19. Nov 3, 2014

### Panphobia

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. Nov 3, 2014

### _N3WTON_

You have to use the chain rule because you have:
$w = f(x,y,z)$ given implicitly by $F(x,y,z,w) = 0$. So you essentially have $F(x,y,z, f(x,y,z)) = 0$. Which is a composite function.