# Find Partial Derivatives of z with Respect to u and v

• -EquinoX-
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
-EquinoX-

## Homework Statement

Find $$\frac{\partial z}{\partial u}$$ and $$\frac{\partial z}{\partial v}$$ using the chain rule.

$$z = \arctan(\frac{x}{y}) , x=u^2+v^2 , y=u^2-v^2$$

## The Attempt at a Solution

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

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

Last edited:
-EquinoX- said:

Are we supposed to guess at what steps are written on your paper?

For one thing d(arctan(x))/dx=1/(1+x^2). I think you missed the square.

oh you're right, I missed the squares, but it is still wrong

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.

well I think the problem here is just from taking the derivative of

$$\frac{u^2+v^2}{u^2-v^2}$$ with respect to u, right. I realize that there should be a - sign in front of it.. am I right?

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.

the best simplification I can think of is:

$$\frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2}$$

Last edited:
Do you mean -4uv/((u^2-v^2)^2+(u^2+v^2)^2)?? I can write that in less space.

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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