# Second partial derivatives

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## Homework Statement

I have an expression for the partial derivative of u with respect to s, which is $$\frac{\partial\,u}{\partial\,s} = \frac{\partial\,u}{\partial\,x}x + \frac{\partial\,u}{\partial\,y}y$$
How do I compute $\frac{\partial^2u}{\partial\,s^2}$ from this?

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Is something missing from the question that is required ?( I have the explicit functions x = x(s,t) and y=y(s,t)). Anyone any ideas?

HallsofIvy
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## Homework Statement

I have an expression for the partial derivative of u with respect to s, which is $$\frac{\partial\,u}{\partial\,s} = \frac{\partial\,u}{\partial\,x}x + \frac{\partial\,u}{\partial\,y}y$$
How do I compute $\frac{\partial^2u}{\partial\,s^2}$ from this?
Use the chain rule:
$$\frac{\partial^2 u}{\partial s^2}= \frac{\partial}{\partial u}\left(\frac{\partial u}{\partial x}x\right)+ \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}y\right)$$

The derivative of anything with respect to s, assuming that s itself is a function of x and y, is
$$\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partal y}\frac{\partial y}{\partial s}$$

Gold Member
Use the chain rule:
$$\frac{\partial^2 u}{\partial s^2}= \frac{\partial}{\partial u}\left(\frac{\partial u}{\partial x}x\right)+ \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}y\right)$$

The derivative of anything with respect to s, assuming that s itself is a function of x and y, is
$$\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partal y}\frac{\partial y}{\partial s}$$
Should that be an s above?
Given this, I have $$\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)x + \frac{\partial}{\partial s}\left(x\right) \frac{\partial u}{\partial x}$$ for the first term.
What do I do from here?

SammyS
Staff Emeritus
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Gold Member

## Homework Statement

I have an expression for the partial derivative of u with respect to s, which is $$\frac{\partial\,u}{\partial\,s} = \frac{\partial\,u}{\partial\,x}x + \frac{\partial\,u}{\partial\,y}y$$
How do I compute $\frac{\partial^2u}{\partial\,s^2}$ from this?
Is something missing from the question that is required ?( I have the explicit functions x = x(s,t) and y=y(s,t)). Anyone any ideas?
Yes, it's important to know that x = x(s,t) and y=y(s,t)

What is the general form of $\displaystyle \frac{\partial\,u}{\partial\,s}\,,\ \ \text{ if } \ \ x = x(s,t)\ \text{ and }\ y=y(s,t)\ ?$

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Yes, it's important to know that x = x(s,t) and y=y(s,t)

What is the general form of $\displaystyle \frac{\partial\,u}{\partial\,s}\,,\ \ \text{ if } \ \ x = x(s,t)\ \text{ and }\ y=y(s,t)\ ?$
Is this not the equation I wrote in my first post?

SammyS
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Is this not the equation I wrote in my first post?
The equation in your first post is for some particular x = x(s,t) and y = y(s,t).

In general, $\displaystyle \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\ .$

This along with the equation in your first post tells you something about $\displaystyle \frac{\partial x}{\partial s}$ and $\displaystyle \frac{\partial y}{\partial s}\ .$

Gold Member
Yes, it tells me that, $$\frac{\partial x}{\partial s} = x \,\,\text{and}\,\,\frac{\partial y}{\partial s} =y$$ I.e the partial derivatives with respect to s are the actual functions.

SammyS
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Yes, it tells me that, $$\frac{\partial x}{\partial s} = x \,\,\text{and}\,\,\frac{\partial y}{\partial s} =y$$
Yes. Now proceed with what you said in post #4 regarding: $\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)\ ,$

$\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)x + \frac{\partial}{\partial s}\left(x\right) \frac{\partial u}{\partial x}\ ,$

which can be written as:

$\displaystyle \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)x + \frac{\partial x}{\partial s}\, \frac{\partial u}{\partial x}\ .$

Gold Member
Yes. Now proceed with what you said in post #4 regarding: $\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)\ ,$

$\displaystyle \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\,x\right)=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)x + \frac{\partial}{\partial s}\left(x\right) \frac{\partial u}{\partial x}\ ,$

which can be written as:

$\displaystyle \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)x + \frac{\partial x}{\partial s}\, \frac{\partial u}{\partial x}\ .$
Where do I go from here?
I have written the above as $$\frac{\partial u}{\partial x} x + \left(\frac{\partial u}{\partial s}\right)\left(\frac{\partial u}{\partial x}\right) x = \frac{\partial u}{\partial x} x \left( 1 + \frac{\partial u}{\partial s}\right)$$

SammyS
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I have written the above as $$\frac{\partial u}{\partial x} x + \left(\frac{\partial u}{\partial s}\right)\left(\frac{\partial u}{\partial x}\right) x = \frac{\partial u}{\partial x} x \left( 1 + \frac{\partial u}{\partial s}\right)$$
Those are not correct .

$\displaystyle \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)x + \frac{\partial x}{\partial s}\, \frac{\partial u}{\partial x}$
$\displaystyle =x\frac{\partial u}{\partial x} + x \left(\frac{\partial^2 u}{\partial s\,\partial x}\right)$

$\displaystyle =x\frac{\partial }{\partial x}\left( u + \frac{\partial u}{\partial s}\right)\,, \ \ \$ but I wouldn't put it into this last form. Perhaps only factor out x.​

Where do I go from here?
Use what HallsofIvy stated in post #3:
...

The derivative of anything with respect to s, assuming that f itself is a function of x and y, is
$$\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$
where he's giving the the result of using the chain rule on, $\displaystyle \frac{\partial f}{\partial s}\ .$

Use $\displaystyle \frac{\partial u}{\partial s}\$ in place of the function, f.

Gold Member
Ok, so $$\frac{\partial}{\partial s} (f) = \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial s} \right) = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial s}\right) \frac{\partial x}{\partial s} + \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial s} \right) \frac{\partial y}{\partial s}$$
But before I computed $$\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right),\,\, \text{and not} \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial s}\right) ?$$

SammyS
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But before I computed $$\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right),\,\, \text{and not} \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial s}\right) ?$$
I failed to notice that you were computing the wrong partial derivative.

What you need is the following:
Ok, so $$\frac{\partial}{\partial s} (f) = \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial s} \right) = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial s}\right) \frac{\partial x}{\partial s} + \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial s} \right) \frac{\partial y}{\partial s}$$
Now, plug-in what you were given for $\displaystyle \frac{\partial\,u}{\partial\,s}$ in the statement of the problem.

The resulting expression will have some occurrences of $\displaystyle \frac{\partial x}{\partial s}$ and$\displaystyle \frac{\partial y}{\partial s}\ .$ Don't forget to change these in accordance with your previous results.