Is my Multivariable Chain Rule Derivation Correct?

[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]

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}}.

 

[professorlucky]

hint

I don't think so since du is not equal to dudu

 

[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:

<br /> \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}}<br />

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]

A second derivative is just the first derivative of what you compute for the first derivative.