# Chained partial derivatives

1. Nov 27, 2013

### Felafel

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

let u=f(x,y) , x=x(s,t), y=y(s,t) and u,x,y$\in C^2$

find:

$\frac{\partial^2u}{\partial s^2}, \frac{\partial^2u}{\partial t^2}, \frac{\partial^2u}{\partial t \partial s}$ as a function of the partial derivatives of f.

i'm not sure i'm using the chain rules correctly, my ideas are a bit confused..

3. The attempt at a solution

1- write f as f(x,y)=f(x(s,t),y(s,t))

$\frac{\partial u}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial f}{\partial x}(x'(s)*x'(s,t))+\frac{\partial f}{\partial y}(y'(s)*y'(s,t))$
Thus, the second derivative:
$\frac{\partial^2 u}{\partial s^2} = \frac{\partial^2 f}{\partial x^2}(x''(s)x'(s,t)+x'(s)x''(s,t))+\frac{\partial^2 f}{\partial y^2}(y''(s)y'(s,t)+y'(s)y''(s,t))$

and then i apply the same for the other two second derivatives.
am i using the right formula?
thank you

2. Nov 27, 2013

### brmath

what is the *?

3. Nov 27, 2013

### Felafel

just the multiplication sign :)

4. Nov 27, 2013

### brmath

This is not quite right. When you write (∂f/∂x)(∂x/∂s) you do not have any further derivatives to take on x. Yes, there is a t in there, but when you are differentiating with respect to s, the t is regarded as a constant. There is no further chain rule to be applied.

Take a specific example to convince yourself of this, say f(x,y) = 2x + 3y; x(s,t) = $s^2 + t^2$ and y(s,t) = st, or anything else you want to make up. Write out f explicitly in terms of s and t. What are your first derivatives with respect to s and t? Now do it again without be explicit, instead using the chain rule.

Hopefully that will help clarify what derivatives go where.

5. Nov 27, 2013

### haruspex

That's fine, but I don't think you can take this any further without information about the functions x(s,t), y(s,t).
That makes no sense. There is no defined function x(s), it's x(s,t). So x'(s) means nothing. x'(s,t) isn't defined either, because it does not indicate whether the derivative is wrt s or t.

6. Nov 27, 2013

### brmath

Hi Haruspex,

One can formally specify the 2nd derivative, even without knowing what the functions are, and that is the question which was asked.

7. Nov 27, 2013

### haruspex

I meant, you can't take the expansion of ∂u/∂s any further.

8. Nov 27, 2013

### brmath

Don't see how one could since we know nothing specific about u. Happy Thanksgiving to you.

B

9. Nov 27, 2013

### haruspex

Right, but in the OP an attempt was made to do so.

10. Nov 28, 2013

### Felafel

thank you. so, apparently i can't be much explicit
i'll go for another try:
$\frac{\partial ^2 f}{\partial s^2}=\frac{\partial f}{\partial x}\left( \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} \right)+\frac{\partial f}{\partial x} \left( \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} \right) = \frac{\partial ^2f}{\partial x^2} \frac{\partial f \partial x}{\partial x \partial s}+\frac{\partial f}{\partial x} \frac{\partial ^2x}{\partial s^2} + \frac{\partial ^2 f}{\partial y^2} \frac{\partial f \partial y}{\partial y \partial s}+\frac{\partial f}{\partial y}\frac{\partial ^2 y}{\partial s^2}$
do i have to leave it this way?

Last edited: Nov 28, 2013
11. Nov 28, 2013

### Felafel

no, sorry, i've just realised i've written nosense.
Here's what i computed:

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

$\frac{\partial ^2 f}{\partial s^2}=\frac{\partial}{\partial s}(\frac{\partial f}{\partial x}\frac{\partial x}{\partial s})+\frac{\partial}{\partial s} (\frac{\partial f}{\partial y}\frac{\partial y}{\partial s})$
using the chain rule again i eventually get
$\frac{\partial ^2 f}{\partial x^2} \frac{\partial ^2 x}{\partial s^2}+\frac{\partial ^2 f}{\partial y \partial x} \frac{\partial y \partial x}{\partial s^2}+\frac{\partial f}{\partial x}\frac{\partial ^2 x}{\partial s^2}$

and without doing all the calculation again for $\frac{\partial ^2 u }{\partial t^2}$ and $\frac{\partial ^2 u}{\partial t \partial s}$ i can just replace the oppurtune s's with t's.

now it should be right..

12. Nov 28, 2013

### haruspex

You're missing some terms. There should be five distinct terms, including components like $\left(\frac{\partial x}{\partial s }\right)^2$.
I've never seen this written: $\frac{\partial y \partial x}{\partial s^2}$. I guess you mean $\frac{\partial x}{\partial s}\frac{\partial y}{\partial s}$

13. Nov 28, 2013

### Felafel

ok, now I've really solved it! i'm too lazy atm to copy it all but i'm finally confident about the result.
thank you a lot :)

14. Nov 28, 2013

### brmath

I also didn't get what you've got here, but glad you sorted it out. Keeping all the different variables straight is the issue, which is no doubt why you were assigned this problem.

Having done that, might I recommend that you tackle some problems with specific f, x,y and see if you can come to the right answers. A book with an answer section would be useful for you.