Implicit differentiation problem

In summary, the conversation is about finding the partial derivatives dz/dx and dz/dy using implicit differentiation in a problem involving arctangent. The conversation includes a discussion about the derivative of arctangent and the use of total and partial differentiation.
  • #1
mr_coffee
1,629
1
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 [Broken]
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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 ?
 
  • #5
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?
 
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  • #6
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.
 

1. What is implicit differentiation and how is it different from explicit differentiation?

Implicit differentiation is a method used to find the derivative of a function that is not written in the form of y = f(x). It is different from explicit differentiation because in explicit differentiation, the dependent variable y is explicitly written as a function of the independent variable x.

2. When should I use implicit differentiation?

Implicit differentiation is typically used when the dependent variable cannot be easily isolated on one side of the equation, or when there are multiple variables and differentiating with respect to one variable would be difficult.

3. How do I solve an implicit differentiation problem?

To solve an implicit differentiation problem, you need to use the chain rule and the product rule, if applicable. You will also need to differentiate both sides of the equation with respect to the independent variable and then solve for the derivative.

4. Can I use implicit differentiation to find higher order derivatives?

Yes, implicit differentiation can be used to find higher order derivatives by repeatedly differentiating both sides of the equation with respect to the independent variable.

5. Are there any limitations to using implicit differentiation?

There are certain limitations to using implicit differentiation. It may not work for all types of equations, and it may be more difficult to solve for the derivative compared to explicit differentiation. Additionally, it may not always provide the most efficient solution for finding derivatives.

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