Find Partial Derivatives of z with Respect to u and v

In summary: Sorry for the confusion.In summary, the chain rule is used to find \frac{\partial z}{\partial u} and \frac{\partial z}{\partial v} for the given equations using the arctan function. The correct derivatives are \frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2} for \frac{\partial z}{\partial u} and \frac{4u^2v}{(u^2-v^2)^2+(u^2+v^2)^2} for \frac{\partial z}{\partial v}. The incorrect answers were due to missing squares in the equations and a missed negative sign. The final equations can
  • #1
-EquinoX-
564
1

Homework Statement


Find [tex] \frac{\partial z}{\partial u} [/tex] and [tex] \frac{\partial z}{\partial v} [/tex] using the chain rule.

[tex] z = \arctan(\frac{x}{y}) , x=u^2+v^2 , y=u^2-v^2 [/tex]

Homework Equations


The Attempt at a Solution



[tex] \frac{\partial z}{\partial u} = \frac{4uv^2}{v^4 - 2u^2v^2 + u^4} * \frac{1}{1+((u^2+v^2)/(u^2-v^2))^2)} [/tex]

[tex] \frac{\partial z}{\partial v} = \frac{4u^2v}{v^4 - 2u^2v^2 + u^4} * \frac{1}{(1+((u^2+v^2)/(u^2-v^2))^2)} [/tex]

why is my answer wrong?
 
Last edited:
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  • #2
-EquinoX- said:
why is my answer wrong?

Are we supposed to guess at what steps are written on your paper? :confused:
 
  • #3
For one thing d(arctan(x))/dx=1/(1+x^2). I think you missed the square.
 
  • #4
oh you're right, I missed the squares, but it is still wrong
 
  • #5
Tom Mattson is right. It would be a lot easier to say what is wrong if you would show what you've done. Just showing a wrong answer and saying 'what did I do wrong?' is more of a puzzle than a question. I don't think you've done much wrong. I get a different sign for dz/du and you could certainly simplify them more.
 
  • #6
well I think the problem here is just from taking the derivative of

[tex] \frac{u^2+v^2}{u^2-v^2} [/tex] with respect to u, right. I realize that there should be a - sign in front of it.. am I right?
 
  • #7
Yes. But like I said you can also simplify those expressions a lot. I don't know if the HW checker requires this or not.
 
  • #8
the best simplification I can think of is:

[tex] \frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2} [/tex]
 
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  • #9
Do you mean -4uv/((u^2-v^2)^2+(u^2+v^2)^2)?? I can write that in less space.
 
  • #10
Yes that's what I meant
 

What is the definition of a partial derivative?

A partial derivative is the rate of change of a function with respect to one of its variables, while holding all other variables constant.

Why do we need to find partial derivatives?

Partial derivatives are important in multivariable calculus because they allow us to analyze how a function changes in multiple directions, rather than just one.

How do you find partial derivatives using the limit definition?

To find the partial derivative of a function f(x,y) with respect to x, we use the limit definition: lim(h->0) (f(x+h,y) - f(x,y)) / h. We then repeat this process for y by holding x constant and taking the limit as k->0.

What are some common notations for partial derivatives?

The most common notations for partial derivatives are: ∂f/∂x, fx, D1f, or ∂1f.

Can you give an example of finding partial derivatives of a function?

Sure, let's find the partial derivatives of f(x,y) = 3x2y + 2xy2. ∂f/∂x = 6xy + 2y2 ∂f/∂y = 3x2 + 4xy

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