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Possible Duplicate:
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.

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    Probably you should say what variables are functions of what other variables. Partial derivatives depend on that information.2012-11-11

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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 $u$.

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    @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.2012-11-12