Partial Differentiation: second partial derivative

[jellicorse]

I am not quite sure how \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)
=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)

comes to \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)...


I would have evaluated this to be the same but without the \frac{\partial z}{\partial x} term. i.e.:

u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)

I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?

 

[Mark44]

I am not quite sure how \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)
=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)

How are z, u, and v related? I can guess, but that really is information you should supply.

comes to \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)...


I would have evaluated this to be the same but without the \frac{\partial z}{\partial x} term. i.e.:

u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)

I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?

 

[jellicorse]

Sorry. z =f (x,y), with

x=\frac{1}{2}(u^2-v^2) and y =uv

 

[Mark44]

Do you understand this part?
\frac{\partial z}{\partial u}=\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)

Or are you having difficulties with this part?
\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )

= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)

 

[jellicorse]

Or are you having difficulties with this part?
\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )

= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)

This is the part that I don't understand. The first part is OK...

 

[Mark44]

Or are you having difficulties with this part?
\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )

= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)

This is the part that I don't understand. The first part is OK...

$$ \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} ) $$
$$ = \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) + \frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

Now use the product rule on each part.

 

[jellicorse]

Thanks a lot, Mark... I am getting closer but have one problem. Having done it several times over, the same thing keeps happening:

Using the product rule (uv)' = u'v + uv' I get one extra term.

I'll type it piece by piece for clarity:


\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) = (uv)'

Gives:

\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})


And


\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} ) = (uv)'

Gives:

\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )


So, in total I get:

\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x})+\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )

=\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})+\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )

 

[Mark44]

You have z = f(x, y), where x = g(u, v) and y = h(u, v). Here x and y are dependent variables (depending on u and v), and u and v are independent variables. Since u and v are independent (are unrelated to each other), $\frac{\partial v}{\partial u} = 0$, so your third term vanishes. By the same reasoning, $\frac{\partial u}{\partial v}$ would also be zero.

Nice job on the LaTeX formatting!

 

[jellicorse]

Oh, I see now. Thanks a lot, it all makes perfect sense now!