# Multivariable Chain Rule

1. Sep 14, 2005

### Icebreaker

Please let me know if I derived this correctly (I did it a while back, and can't find the notebook):

$$v(x,y)=u(r(x,y),s(x,y))$$

(derivations)

At some point I come across this:

$$\frac{\partial}{\partial x} \frac{\partial u}{\partial r}$$

which I wrote as

$$\frac{\partial^2 u}{\partial r^2} \frac{\partial r}{\partial x}$$

Is it right?

2. Sep 14, 2005

### HallsofIvy

Since u depends on r and s, and r and s are both functions of x, you are going to have to take into account the dependence of x on s.

For any function φ(x,y), $$\frac{\partial \phi}{\partial x}= \frac{\partial \phi}{\partial r}}\frac{\partial r}{\partial x}}+ \frac{\partial \phi}{\partial s}}\frac{\partial s}{\partial x}}$$.

Now put $$\frac{\partial u}{\partial r}}$$ in place of φ
You get $$\frac{\partial^2 \phi}{\partial r^2}}\frac{\partial r}{\partial x}}+ \frac{\partial^2 \phi}{\partial r\partial s}}\frac{\partial s}{\partial x}}$$.

Last edited by a moderator: Sep 14, 2005
3. Sep 17, 2005

### professorlucky

hint

I dont think so since du is not equal to dudu

4. Sep 17, 2005

### Benny

If you only need to differentiate once then the following procedure might come in handy. Say for example you have f(x,y,z) where x,y,z are functions of s and t and you needed to the find the partial derivative of f wrts. What you could do is draw a tree diagram.

You standard with f and the top and draw out three branches, one to each of x, y and z. In a similar manner you do the same with each of x, y and z. That is, draw two branches from each of x, y and z to s and t. You should get a pyramid like diagram after you do this. To find $$\frac{{\partial f}}{{\partial s}}$$ all you would need to do is draw the appropriate path/s. In other words you just go from f along any path where the end point is s. You then 'sum your paths.' The only restriction is that you keep going downward.

In this case, you would get:

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

someone please correct me if my answer is incorrect because I didn't write it down on paper.

5. Sep 18, 2005

### Icebreaker

The chain rule for the first derivative I can handle. It's when you have the second and third derivatives that I can't follow which function is related to what.

6. Sep 18, 2005

### Hurkyl

Staff Emeritus
A second derivative is just the first derivative of what you compute for the first derivative. :tongue2: