How do you find partial derivatives for tan(x/y)?

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Homework Help Overview

The discussion revolves around finding the first partial derivatives of the function tan(x/y) with respect to the variables x and y. Participants explore the application of partial differentiation and the use of the chain rule in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to differentiate tan(x/y) and questions how to apply the product rule in the context of partial derivatives. Some participants provide examples of partial derivatives with different functions, while others suggest methods for approaching the differentiation.

Discussion Status

Some participants have shared their attempts at calculating the partial derivatives, with one confirming the correctness of another's results. There is ongoing exploration regarding the application of similar techniques to a different function involving logarithms, with some participants expressing uncertainty about notation and the validity of certain expressions.

Contextual Notes

There are indications of confusion regarding notation and the interpretation of expressions related to the logarithmic function, as well as a request for clarification on specific parts of the problem.

elle
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Hi, can anyone help me with the following differentiation question?

Find first partial derivatives w.r.t to x and y for:

tan (x/y)

Can anyone offer any help on how to approach this question? I know when you differentiate (x/y) using the product rule, that you have to differentiate y implicitly...but how does that work when using partial derivatives? :confused:

Thanks for taking time to read :smile:
 
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In partial derivatives, you treat all variables as constants (you know, the a, b, c)... except for the one you are differentiating with respect to.

For example:

f(x,y,z) = 2xyz + 3x + 4y + 5z
df/dz = 2xy + 5
df/dx = 2yz + 3
df/dy = 2xz + 4
 
Ok, I've given it a go :rolleyes: but I'm not sure if its right...

f (x,y) = tan (x/y)

Let f(x,y) = tan (u) where u = (x/y) = xy^(-1) *** Not sure if I've used the right method for this part

f1 (w.r.t x) = (1/y) sec^2 (x/y)

f2 (w.r.t y) = -(x/y^2) sec^2 (x/y)

Hmm can someone let me know if its right? :rolleyes:
 
Yes, you have gotten the correct answers.
 
Thank you! Glad to know I am on the right track :biggrin:

If I had to use partial differentiation for (1/2) ln (x^2 + y^2), can I just use the same technique (chain rule) and:

Let (1/2) ln (x^2 + y^2) = 1/2 ln (u) where u = x^2 + y^2

Then:

f1 (w.r.t x) = x ln (x^2 + y^2)

f2 (w.r.t y) = y ln (x^2 + y^2) ?
 
Certainly you can do that.
 
ok thanks! There is just one last part of da question regarding the logs which I don't understand.

Show that x f(1) + y (f2) = 1? :confused:

I've started with:

x^2 ln (x^2 + y^2) + y^2 ln (x^2 + y^2)

so how do I show that it is equal to 1?
 
elle said:
ok thanks! There is just one last part of da question regarding the logs which I don't understand.

Show that x f(1) + y (f2) = 1? :confused:

I've started with:

x^2 ln (x^2 + y^2) + y^2 ln (x^2 + y^2)

so how do I show that it is equal to 1?

I don't understand this. Is "x f(1)+ y(f2)= 1" supposed to be
xfx+ yfy= 1?

Are you saying that f(x,y)= x2ln(x2+y2)+ y2ln(x2+y2)?

If those are what you mean, then you can't show that
xfx+ yfy= 1, it's not true.
 
elle said:
ok thanks! There is just one last part of da question regarding the logs which I don't understand.

Show that x f(1) + y (f2) = 1? :confused:

I've started with:

x^2 ln (x^2 + y^2) + y^2 ln (x^2 + y^2)

so how do I show that it is equal to 1?
LEARN PROPER NOTATION!

Now, f(1) as you call it should be [tex]\frac{\partial{f}}{\partial{x}}=\frac{x}{x^{2}+y^{2}}[/tex] and f(2) is really [tex]\frac{\partial{f}}{\partial{y}}=\frac{y}{x^{2}+y^{2}}[/tex]
when [tex]f(x,y)=\frac{1}{2}ln(x^{2}+y^{2})[/tex]
 
  • #10
Yes it is, I don't know how to do the superscript x you see :blushing:

Erm I've attached the question here, though its suppose to be 'u' not 'f'

http://tinypic.com/eg5mhi.jpg"
 
Last edited by a moderator:
  • #11
elle said:
Yes it is, I don't know how to do the superscript x you see :blushing:

Erm I've attached the question here, though its suppose to be 'u' not 'f'

http://tinypic.com/eg5mhi.jpg"
Just click on the Latex images to see how the code is for the various symbols.
 
Last edited by a moderator:

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