Partial Derivatives: Finding dz/dx w/o z=u^v

[Firepanda]

http://img132.imageshack.us/img132/5736/wathh1.jpg

I think to find dz/dx (d = delta) first of all is by

dz/dx = (dz/du)(du/dx) + (dz/dv)(dv/dx)

But how do I find dz/du and dz/dv for this?

I only have 1 example that resembles this and it had z defined as a function as z = u^v, but here I don't have that, so how can I find it?

Thanks

 

[rocomath]

find dz/du * du/dx - take the derivative of u with respects to x

 

[Firepanda]

Sorry I don't understand, I'm trying to find dz/du first then dz/dx.

If I take the derivative of u with respect to x and hold y constant, then i get the partial derivative of du/dx. Don't I?

 

[DavidWhitbeck]

I don't think you mean delta. I think that you're writing the chain rule and meant to write partial derivatives and the d's are $$\partial$$ right?

I'll give you an example:

$$z = uv, u=\cos (x+y), v=\sin (x+y)$$
then clearly $$z=\frac{1}{2}\sin2(x+y) \Rightarrow \frac{\partial z}{\partial x} = \cos2(x+y)$$

But now let's use your formula and see if we get the same thing:
$$\frac{\partial z}{\partial u}=v=\sin(x+y),\frac{\partial u}{\partial x}=-\sin(x+y)$$
$$\frac{\partial z}{\partial v}=u=\cos(x+y),\frac{\partial v}{\partial x}=\cos(x+y)$$

So that
$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \Rightarrow$$
$$\frac{\partial z}{\partial x}=-\sin(x+y)\sin(x+y)+\cos(x+y)\cos(x+y)$$
$$\frac{\partial z}{\partial x}=\cos2(x+y)$$

And hey it's the same thing! So in that example all the derivatives are partial derivatives, and you keep the other variables constant when you evaluate them.

 

[Firepanda]

Ah sorry ye I thought the curly d was a delta :P

 

[Firepanda]

Thanks for your reply, so am I right in assuming dz/du = v and dz/dv = u?

I didn't know that, thanks.

 

[DavidWhitbeck]

Thanks for your reply, so am I right in assuming dz/du = v and dz/dv = u?

I didn't know that, thanks.

Not in general, just in that example. For instance if you have $$z=u^2v^3$$ then $$\frac{\partial z}{\partial u}=2uv^3$$ and $$\frac{\partial z}{\partial v}=3uv^2$$.

 

[Firepanda]

ok well in my question

http://img132.imageshack.us/img132/5736/wathh1.jpg

I plan on finding dz/du and dz/dv, but how do I go about this?

I don't have a z = .. like you are using in your examples unless I am misunderstanding the notation in the question.

I then plan on spending a while figuring out where to go from there :)

Edit: It says z = f(x,y) but I'm not sure what this means exactly.

 

[Firepanda]

:/ I just found more examples in a textbook, they all have f(x,y) equal to something, here I do not, is it a mistake with the question?

Its only worth 4 marks and it's taking me forever to do.

 

[Hootenanny]

I'm assuming that your function f is complex valued and therefore may be written,

$$z = f(x,y) = u + i v = x+ay + i\left(x-ay\right)$$

Therefore,

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + i\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$

$$\Rightarrow z_x = \frac{\partial}{\partial x}\left(x+ay\right) + i\frac{\partial}{\partial x}\left(x-ay\right)$$

An similarly for partial derivatives with respect to y. Does that help?

 

[Firepanda]

Yes thankyou, I'll work on from there :)

 

[rootX]

:/ I just found more examples in a textbook, they all have f(x,y) equal to something, here I do not, is it a mistake with the question?

Its only worth 4 marks and it's taking me forever to do.

I am using Glyn James, and I had like 20 these like questions! :D (almost in every single exercise related to partial ...), and also we always get this type of question on the exam:

Here's two good questions:
1)
u = xy
v = x-y
z = f(u,v)

show :
x.dz/dx - y.dz/dy = (x-y)* (dz/du)

2)(little harder)
u = x+ay
v = x+by
z = f(u,v)

show
9 (d^2f/dx^2) - 9 (d^2f/dx.dy) + 2 (d^2f/dy^2) = 0
into
d^2f / du.dv = 0Main strategy to solve these questions: draw a tree with vars!

 

[rootX]

Hootenanny, your solution does not any sense to me >>

Here's how I think it should be done:
z = f(x,y)
x = u+v
y = (u-v)/a

dz/du = dz/dx . dx/du + dz/dy. dy/du
-Applying multiplication rule ..
d^2z / dv.du = d/dv (dz/dx . dx/du ) + ...
= [d^z/dx^2 . (dx/du)^2 + d^2z /dx.dy * dx/du * dy/du ]+ dz/dx * dx/ du.dv * dv/dx ...

wow, it so messy but it should go like that if not necessarily exactly like above
and you final answer may have some deltas in it

 

[Hootenanny]

Hootenanny, your solution does not any sense to me >>

Here's how I think it should be done:
z = f(x,y)
x = u+v
y = (u-v)/a

dz/du = dz/dx . dx/du + dz/dy. dy/du
-Applying multiplication rule ..
d^2z / dv.du = d/dv (dz/dx . dx/du ) + ...
= [d^z/dx^2 . (dx/du)^2 + d^2z /dx.dy * dx/du * dy/du ]+ dz/dx * dx/ du.dv * dv/dx ...

wow, it so messy but it should go like that if not necessarily exactly like above
and you final answer may have some deltas in it

z is usually used to denote complex variables, hence I assumed that f(x,y) was complex valued. If f is indeed complex valued then u and v simply represent the real and imaginary parts of the function and the function can be differentiated as I have shown above.

 

[rootX]

ook thnx.
They never told/made us think that way><
[But thinking either way shouldn't give different answer, right?]

 

[Lyn]

I'm assuming that your function f is complex valued and therefore may be written,

$$z = f(x,y) = u + i v = x+ay + i\left(x-ay\right)$$

Therefore,

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + i\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$

$$\Rightarrow z_x = \frac{\partial}{\partial x}\left(x+ay\right) + i\frac{\partial}{\partial x}\left(x-ay\right)$$

An similarly for partial derivatives with respect to y. Does that help?

Hi i am struggling too with this same question, http://img132.imageshack.us/img132/5736/wathh1.jpg

I have tried to follow what you have said and i get my (d's are partial derivatives) dz/dx = 2. I am unsure if this is correct because just like the problem the other person had with this question, i don't have a z=... (eg z=xy)