Implicit differentiation problem

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Homework Help Overview

The discussion revolves around an implicit differentiation problem from calculus, specifically focusing on finding the derivatives dz/dx and dz/dy for the equation x - z = arctan(yz).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the differentiation of the given equation and raise questions about the correct application of derivatives, particularly regarding the arctangent function. There is also discussion about the interpretation of total versus partial derivatives.

Discussion Status

The conversation includes various attempts to differentiate the equation and clarify misunderstandings about derivative notation. Some participants provide guidance on manipulating the derivatives, while others express confusion about the notation used and its implications.

Contextual Notes

There is mention of potential misinterpretation between total and partial differentiation, which may affect the understanding of the problem. The original poster's use of notation is questioned, indicating a possible source of confusion in the discussion.

mr_coffee
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hello everyone I'm stuck! anyone have any ideas?
I'm suppose to find dz/dx and dz/dy with implicit differentation. This is calc III!
http://img221.imageshack.us/img221/4000/lastscan4ou.jpg
 
Last edited by a moderator:
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Is the question "x - z = arctan(yz). Find dz/dx. Find dz/dy." ?

Differentiate throughout: dx - dz = (ydz+zdy)arctan'(yz).

Then manipulate into dz = A + Bdx where A and B are functions of x, y and z. Divide by dx to get dz/dx = A/dx + B.

Similar for dz/dy.
 
One problem you have is that you have the wrong derivative for arctangent!

The derivative of arctan(x) is [tex]\frac{1}{1+ x^2}[/tex]

if x-z= arctan(yz) then, writing zx and zy for the derivatives of z with respect to x and y respectively, we have
[tex]1-z_x= \frac{yz_x}{1+ y^2z^}[/tex]
which you can solve for zx and
[tex]-z_y= \frac{z+ yz_y}{1+ y^2z^2}[/tex]
which you can solve for zy.
 
One thing I couldn't understand here was, what happened to dy/dx and dx/dy.

We have,

x - z = arctan(yz)

differentiating wrt x,

1 - dz/dx = d/dx{arctan(s)}, where s = yz
1 - dz/dx = d/ds{arctan(s)}.ds/dx
1 - dz/dx = 1/(1 + s²) * (y.dz/dx + z.dy/dx)
1 - dz/dx = (y.dz/dx + z.dy/dx) / (1 + y²z²)

Adopting HallsofIvy's notation,

1 - zx = (yzx + zyx)/(1 + y²z²)

What have I missed out ?
 
Fermat said:
One thing I couldn't understand here was, what happened to dy/dx and dx/dy.
Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "[itex]\partial[/itex]" notation?
 
Last edited:
EnumaElish said:
Could it be that the OP used total diff. "d" notation but HallsOfIvy interpreted it as partial diff. "[itex]\partial[/itex]" notation?
That explains it.
Thanks.
 

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