What $\partial^2 u / \partial s^2$ is equal to?

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How does partial derivative work?

If $\frac{\partial u}{\partial s}$ is equal to $$e^s \cos(t) \frac{\partial u}{\partial x} + e^s \sin(t) \frac{\partial u}{\partial y},$$ what $\frac{\partial^2 u}{\partial d s^2}$ is equal to?

What expression are we suppose to get? I have been trying to figure out what to do for a hour, but I am quite lost. Doesn't the second partial derivative of $s$, give the same thing?

Any help would be appreciated. Thanks.

$\partial ds^2$ looks like a typo. There shouldn't be both a $\partial$ and a $d$. — hmakholm left over Monica, Nov 11 '12 at 19:16
Probably you should say what variables are functions of what other variables. Partial derivatives depend on that information. — GEdgar, Nov 11 '12 at 19:56

Thomas · Answer 1 · 2012-11-11T19:45:52.813

0

We have $$ \frac{\partial^2 u}{\partial s^2} = \frac{\partial}{\partial s} (\partial u / \partial s). $$ In your case $cos(t)$ is considered a constant. We have $u$ a function of $s$, and likewise the derivatives of $u$.

So $$\begin{align} \frac{\partial^2 u}{\partial s^2} &= \frac{\partial}{\partial s} (\partial u / \partial s) \\ &= \frac{\partial}{\partial s} (e^s cos(t) \partial u / \partial x + e^s sin(t) \partial u / \partial y)\\ &= \left[(e^s cos(t) \partial u / \partial x)+ (e^s cos(t) \frac{\partial^2 u}{ \partial s\partial x})\right] + \left[(e^s sin(t) \partial u / \partial y) + (e^s sin(t) \frac{\partial^2 u}{\partial s\partial y})\right] \end{align} $$ Note that we have used the product rule here since you for example have the product of $e^s$ with the partial derivative of $u$.

edited Nov 11 '12 at 19:45

answered Nov 11 '12 at 19:21

Thomas

44,491

what did you do? how come you have 3 terms instead of 2? – question Nov 11 '12 at 19:23
err it should be s instead of 2 – question Nov 11 '12 at 19:25
@Matt. The first term is a product of two functions that depend on $s$, so using the product rule... – Thomas Nov 11 '12 at 19:25
i don't understand what \partial / partial s is suppose to do... is that supposed to be the partial derivative of s? – question Nov 11 '12 at 19:26
but it's the second derivative of s so the trigonometric terms should be considered to be constant variables and not functions. we have f(x) but not g(x). am i wrong? – question Nov 11 '12 at 19:28
btw i made an error, it should be e^s – question Nov 11 '12 at 19:30
@Matthew: I will update – Thomas Nov 11 '12 at 19:40
@Thomas Judging from this question, $x$ and $y$ are functions of $s$. – Pragabhava Nov 11 '12 at 19:41
@Pragabhava: I am a bit confused about this. I will let my answer stand as I feel that it best answers this current question. – Thomas Nov 11 '12 at 19:48
i am even more confused. i couldn't find any video tutorial on this. – question Nov 12 '12 at 00:19
@Matthew: I understand that you are confused. And, as I mentioned elsewhere, I really do think that you just need to study exactly what a partial derivative is and how it is different from a total derivative. – Thomas Nov 12 '12 at 00:23

What $\partial^2 u / \partial s^2$ is equal to?

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