# Partial derivative

## 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

You have to use the chain rule..

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

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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)?

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

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

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

Chet

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..

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.

Chestermiller
Mentor
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

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

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

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 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:
$\frac{\partial F}{\partial x} * 0 = 0$...there are some terms in your answer that should not be there...

andrewkirk
Homework Helper
Gold Member
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
$$\frac{\partial z}{\partial z}=1$$

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.

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..

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:
$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$

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.

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.

andrewkirk
Homework Helper
Gold Member
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

$$\frac{dw}{dz}= \frac{\partial w}{\partial x}\frac{dx}{dz}+ \frac{\partial w}{\partial y}\frac{dy}{dz}+ \frac{\partial w}{\partial z}\frac{dz}{dz}=0+0+1\times\frac{\partial w}{\partial z}$$

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

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

Chet