# Multivariable Chain Rule

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?

HallsofIvy
Homework Helper
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 $$\frac{\partial^2 \phi}{\partial r^2}}\frac{\partial r}{\partial x}}+ \frac{\partial^2 \phi}{\partial r\partial s}}\frac{\partial s}{\partial x}}$$.

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hint

I dont think so since du is not equal to dudu

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. 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.

Hurkyl
Staff Emeritus