Derivative of sqrt(x^2+y^2)=2arctg(y/x)

  • Thread starter Thread starter Penultimate
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around finding the derivative of the equation \(\sqrt{x^2+y^2}=2\arctan\left(\frac{y}{x}\right)\), with participants exploring the context of implicit differentiation and the roles of the variables involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants question whether to find partial derivatives or to assume \(y\) is a function of \(x\) for \(dy/dx\). There are discussions about the application of the chain rule and implicit differentiation, with some participants expressing uncertainty about their attempts and the requirements of the problem.

Discussion Status

The discussion is ongoing, with various participants attempting to clarify the problem's requirements and expressing their confusion about the differentiation process. Some guidance has been offered regarding the chain rule and the nature of the derivative being sought, but no consensus has been reached on the approach.

Contextual Notes

There is a lack of clarity regarding whether the derivative should be taken with respect to \(x\) or if partial derivatives are needed, which has led to differing interpretations among participants.

Penultimate
Messages
26
Reaction score
0
[tex]\sqrt{x^2+y^2}[/tex]=2arctg[tex]\frac{y}{x}[/tex]

Dont know where to start with this one either.
 
Physics news on Phys.org
Penultimate said:
[tex]\sqrt{x^2+y^2}[/tex]=2arctg[tex]\frac{y}{x}[/tex]

Dont know where to start with this one either.
Simply saying "Do the derivate" (or derivative) isn't sufficient. The derivatie with respect to what variable? Is the problem to find partial derivatives of that with respect to x and y or are we to assume that y is a function of x and you wish to find dy/dx?

In either case, this is the same as
[tex](x^2+ y^2)^{1/2}= 2 arctan(\frac{y}{x})[/tex]
If you want to find dy/dx, can you differentiate [tex](x^2+ y^2)^{1/2}[/tex] with respect to x? What about 2 arctan(y/x)? Of course, you will need to use the chain rule. If f is a function of y, which is a function of x, then df/dx= (df/dy)(dy/dx).
 
Last edited by a moderator:
The exercise requires the differential of the expression. I think that is dy/dx.
 
So y is a function of x and the derivative is with respect to x. Attempt the problem first and show where you get stuck.

Recall the chain rule: [f(g(x))]' = f'(g(x)) g'(x). In this case y = g(x)
 
I know is something with derivating twice but i am not figuring anything out.
 
Have you learned the chain rule?
 
Penultimate said:
I know is something with derivating twice but i am not figuring anything out.
So far you haven't shown that you are even trying. What have you done?
 
Penultimate said:
The exercise requires the differential of the expression. I think that is dy/dx.
No, the exercise asks for the derivative, dy/dx, not the differential.
 
Penultimate said:
I know is something with derivating twice but i am not figuring anything out.
No, you don't need to differentiate twice. Once will be enough to find the derivative, but you will need to use implicit differentiation.
 

Similar threads

Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
2K