Partial Differentiation query

Brussels Sprout

Hi,

I’m doing some partial differentiation and I’m just confused about a certain derivation.

z is a function of two independent variables x and y

z =f(x,y)

we can say that

[img]http://latex.codecogs.com/gif.latex?\delta z = \frac{\partial z} {\partial x} \delta x + \frac{\partial z} {\partial y} \delta y[/img]

Dividing through by delta x:

[img]http://latex.codecogs.com/gif.latex?\frac{\delta z} {\delta x} = \frac{\partial z} {\partial x} + \frac{\partial z} {\partial y} \frac{\delta y} {\delta x}[/img]

Then if we let delta x approach zero we get:

[img]http://latex.codecogs.com/gif.latex?\frac{d z} {dx} = \frac{\partial z} {\partial x} + \frac{\partial z} {\partial y} \frac{d y} {dx}[/img]


Now if z=f(x,y) and x and y are themselves functions of two new independent u and v.
We can start with the same equation

[img]http://latex.codecogs.com/gif.latex?\delta z = \frac{\partial z} {\partial x} \delta x + \frac{\partial z} {\partial y} \delta y[/img]

Now if we divide across by delta u we get

[img]http://latex.codecogs.com/gif.latex?\frac{\delta z} {\delta u} = \frac{\partial z} {\partial x} \frac{\delta x} {\delta u} + \frac{\partial z} {\partial y} \frac{\delta y} {\delta u}[/img]

Then as we let delta u approach zero we get:

[img]http://latex.codecogs.com/gif.latex?\frac{\partial z} {\partial u} = \frac{\partial z} {\partial x} \frac{\partial x} {\partial u} + \frac{\partial z} {\partial y} \frac{\partial y} {\partial u}[/img]

My question is why is it that the expression on the left hand side of the equation becomes dz/dx in the first incident and partial derivative of z with respect to u in the second case? I have previously accepted this but I’ve run into trouble later on and I just want to clear up the fundamentals.

Michael Collins

Brussels Sprout wrote: »

Hi,

I’m doing some partial differentiation and I’m just confused about a certain derivation.

z is a function of two independent variables x and y

z =f(x,y)

we can say that

[img]http://latex.codecogs.com/gif.latex?\delta z = \frac{\partial z} {\partial x} \delta x + \frac{\partial z} {\partial y} \delta y[/img]

Dividing through by delta x:

[img]http://latex.codecogs.com/gif.latex?\frac{\delta z} {\delta x} = \frac{\partial z} {\partial x} + \frac{\partial z} {\partial y} \frac{\delta y} {\delta x}[/img]

Then if we let delta x approach zero we get:

[img]http://latex.codecogs.com/gif.latex?\frac{d z} {dx} = \frac{\partial z} {\partial x} + \frac{\partial z} {\partial y} \frac{d y} {dx}[/img]


Now if z=f(x,y) and x and y are themselves functions of two new independent u and v.
We can start with the same equation

[img]http://latex.codecogs.com/gif.latex?\delta z = \frac{\partial z} {\partial x} \delta x + \frac{\partial z} {\partial y} \delta y[/img]

Now if we divide across by delta u we get

[img]http://latex.codecogs.com/gif.latex?\frac{\delta z} {\delta u} = \frac{\partial z} {\partial x} \frac{\delta x} {\delta u} + \frac{\partial z} {\partial y} \frac{\delta y} {\delta u}[/img]

Then as we let delta u approach zero we get:

[img]http://latex.codecogs.com/gif.latex?\frac{\partial z} {\partial u} = \frac{\partial z} {\partial x} \frac{\partial x} {\partial u} + \frac{\partial z} {\partial y} \frac{\partial y} {\partial u}[/img]

My question is why is it that the expression on the left hand side of the equation becomes dz/dx in the first incident and partial derivative of z with respect to u in the second case? I have previously accepted this but I’ve run into trouble later on and I just want to clear up the fundamentals.

In the first one you have the 'total derivative', not quite the same as the partial derivative. Here you are not assuming that y is constant when x varies. If you were, the second factor in the last term would = 0 and you'd be left with just the partial derivative of z with respect to x.

The second one is just the bog standard parial derivative using the chain rule. Here you are assuming that v is constant while u varies. Obviously u can cause changes to z via x=x(u,v) and y=y(u,v), hence these must be included, but v itself never appears.

Brussels Sprout

Michael Collins wrote: »

In the first one you have the 'total derivative', not quite the same as the partial derivative. Here you are not assuming that y is constant when x varies. If you were, the second factor in the last term would = 0 and you'd be left with just the partial derivative of z with respect to x.

The second one is just the bog standard parial derivative using the chain rule. Here you are assuming that v is constant while u varies. Obviously u can cause changes to z via x=x(u,v) and y=y(u,v), hence these must be included, but v itself never appears.

Thank you but this answer deals more with the final results than the process by which they were reached. It is the process that is causing my confusion. In particular the line that says that as delta x or delta x->0 we then end up with either a total or partial derivative on the LHS.

I understand what you're saying about if the first one led to a partial derivative then we'd just end up with the same thing on both sides of the equation but it seems almost arbitrary to have the results based on the outcome (if you know what I mean). It's almost like the second one is a partial derivative because it can be.

Brussels Sprout

Ok I think I have it. In the second set of equations as the delta u -> 0 you can either choose to treat the LHS as a total derivative (in which case the v's also vary and the RHS of the equation gets a bit more complicated) or you can choose to treat the LHS as the the partial derivative in which case the v's are constant and we end up with the equation as shown.

That may sound like I'm just rehashing your answer but the key concept for me was that you have a choice as to what you want the LHS to become. I was thinking that as delta u -> 0 the LHS had to become a total derivative or a partial derivative (ie one was right and one was wrong) and I couldn't understand how you chose which was the correct one. I didn't realise that you can make it either as long as you follow the appropriate rules on the RHS.

Does it sound like I've got it?

Michael Collins

Brussels Sprout wrote: »

Ok I think I have it. In the second set of equations as the delta u -> 0 you can either choose to treat the LHS as a total derivative (in which case the v's also vary and the RHS of the equation gets a bit more complicated) or you can choose to treat the LHS as the the partial derivative in which case the v's are constant and we end up with the equation as shown.

That may sound like I'm just rehashing your answer but the key concept for me was that you have a choice as to what you want the LHS to become. I was thinking that as delta u -> 0 the LHS had to become a total derivative or a partial derivative (ie one was right and one was wrong) and I couldn't understand how you chose which was the correct one. I didn't realise that you can make it either as long as you follow the appropriate rules on the RHS.

Does it sound like I've got it?

Yes exactly!

It comes down to your choice for [latex] \frac{\delta x} {\delta u} [/latex] and [latex] \frac{\delta y} {\delta u} [/latex] in the second case. We should really write these as

[latex] \displaystyle \frac{\delta x(u,v)} {\delta u} [/latex]

and

[latex] \displaystyle \frac{\delta y(u,v)} {\delta u} [/latex]

Now when you take the limit of these two, if you choose v to be constant, that's the definition of partial derivative - which is what you wrote. If not, and v actually depends on u (in some way), then you need the total derivative here, and so

[latex] \displaystyle \frac{\delta x(u,v)} {\delta u} = \frac{\partial x}{\partial u} + \frac{\partial x}{\partial v}\frac{\delta v}{\delta u} [/latex]

and

[latex] \displaystyle \frac{\delta y(u,v)} {\delta u} = \frac{\partial y}{\partial u} + \frac{\partial y}{\partial v}\frac{\delta v}{\delta u} [/latex]

putting this all together you get

[latex] \displaystyle \frac{\delta z}{\delta u} = \frac{\partial z}{\partial x}\left( \frac{\partial x}{\partial u} + \frac{\partial y}{\partial v}\frac{\delta v}{\delta u} \right) + \frac{\partial z}{\partial y} \left( \frac{\partial y}{\partial u} + \frac{\partial y}{\partial v}\frac{\delta v}{\delta u} \right) [/latex]

finally taking the limit as [latex] \delta u \to 0 [/latex] we get:

[latex] \displaystyle \frac{dz}{du} = \frac{\partial z}{\partial x}\left( \frac{\partial x}{\partial u} + \frac{\partial y}{\partial v}\frac{dv}{du} \right) + \frac{\partial z}{\partial y} \left( \frac{\partial y}{\partial u} + \frac{\partial y}{\partial v}\frac{dv}{du} \right) [/latex]

which is just the total derivative of z with respect to u (and could be obtained directly, much quicker!).

Brussels Sprout

Great stuff-I ended up doing the same as you there-breaking it down to it's basic components and going through it step by step and I ended up with that same big equation which simplified nicely in the end to

[img]http://latex.codecogs.com/gif.latex?\frac{d z} {du} = \frac{\partial z} {\partial u} + \frac{\partial z} {\partial v} \frac{d v} {du}[/img]

Thanks for the help!


(I think you might have copied and pasted your (delta y / delta u) bit twice there in the last two equations btw)

Michael Collins

Brussels Sprout wrote: »

...Thanks for the help!

(I think you might have copied and pasted your (delta y / delta u) bit twice there in the last two equations btw)

No bother.

Ack, you're right. Few other mistakes there too but all fixed now I hope.