Implicit differentiation of many variables

In summary: It is a complete and accurate statement of the problem. In summary, a conversation took place about finding the partial derivatives of a given function z to demonstrate an equality. The speaker had some questions about how the derivatives were calculated and suggested that there may be a mistake in the given solution. They proposed a different function for z and calculated the derivatives, which resulted in a true equation. However, the expert pointed out that the solution may have used a different function for z, leading to the discrepancy. The conversation then shifted to discussing how to differentiate a function with multiple variables.
  • #1
jonjacson
447
38

Homework Statement



For the given function z to demonstrate the equality:

10wl0qp.jpg
[/B]As you see I show the solution provided by the book, but I have some questions on this.

I don't understand how the partial derivative of z respect to x or y has been calculated.

Do you think this is correct?

I think this is a giant errata, I guess the function z is not given implicitly and it simply is:

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

The partial derivatives are calculated normally:

∂z/∂x= 2 * x/(x^2 + y^2)

Similar for y, and with this it is straighforward to demonstrate the equality.

What do you think? There are two options:

1.- Or the statement and solution of the given problem is correct---> In that case I don't understand anything. Could you explain how to get the partial derivatives?

2.- Or there is a giant errata, z is not given implicitly and the calculation is easy.

And forgeting this problem I was wondering in case I found an equation with z given implicitly like:

z^2 = x * z + y * z^3

How would we differenciate this equation?

As we have many variables we should choose which are maintained constant and which are changing. Suppose we differenciate this expression considering x is changing, y is constant but z obviously changes, due to the changes in x.

The receipt is changing x for x+dx, z changes to z+dz and y doesn't change at all. I get:

(z+dz)2 - z2 = ( (x+dx) * (z+dz) + y * (z+dz)3 ) - (x * z + y * z3)

After neglecting diferentials of order two and three I get:

dz = dx * (z dx / 2x - x -3 y z^2)

But this differential arose because there was a change on x, so I should call it dzx, then I should do the same calculation for dzy and the total differential of the function z should be:

dz = dzx + dz y

Is this correct?

 
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  • #2
jonjacson said:

Homework Statement



For the given function z to demonstrate the equality:

10wl0qp.jpg
[/B]
Is there information missing from the image above, especially in the upper right corner?
 
  • #3
jonjacson said:

Homework Statement

[/b]
I don't understand how the partial derivative of z respect to x or y has been calculated.

Do you think this is correct?

I think this is a giant errata, I guess the function z is not given implicitly and it simply is:

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

The partial derivatives are calculated normally:

∂z/∂x= 2 * x/(x^2 + y^2)
This result doesn't match what's given in the solution, so why do you think your guess for ##z## is correct?

The solution is definitely wrong for the given ##z##, but your guess is wrong too. It looks like the solution used ##z = \ln (x^2+xy+y^2)##.
 
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  • #4
vela said:
This result doesn't match what's given in the solution, so why do you think your guess for ##z## is correct?

The solution is definitely wrong for the given ##z##, but your guess is wrong too. It looks like the solution used ##z = \ln (x^2+xy+y^2)##.

Thanks for your answer.

But with my guess I find:

∂z/∂x= 2x/(x^2 + y^2)

∂z/∂y= 2y/(x^2 + y^2)

So if I substitute in the equation I get:

x * (2x/(x^2 + y^2)) + y * (2y/(x^2 + y^2)) = 2

In the denominator we have the same functions, so we can simply sum the numerators to get:

(2 x^2 + 2 y^2 )/ (x^2 + y^2) = 2

And this equation is true. What am I doing wrong?

Mark44 said:
Is there information missing from the image above, especially in the upper right corner?

No, there is nothing.
 

Related to Implicit differentiation of many variables

What is implicit differentiation of many variables?

Implicit differentiation of many variables is a mathematical technique used to find the derivative of a function with multiple variables. It is used when the equation of a function is given implicitly, meaning it is not explicitly written as y = f(x).

How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used when the equation of a function is given explicitly, meaning it is written as y = f(x). Implicit differentiation is used when the equation is given implicitly, meaning it is not explicitly written as y = f(x).

What are the steps for performing implicit differentiation of many variables?

The steps for performing implicit differentiation of many variables are as follows:

  1. Take the derivative of each term in the equation with respect to the variable of interest.
  2. Combine like terms and move all terms containing the derivative of the variable to one side of the equation.
  3. Solve for the derivative of the variable by isolating it on one side of the equation.

What is an example of using implicit differentiation of many variables?

An example of using implicit differentiation of many variables is finding the derivative of the equation x^2 + y^2 = 25 with respect to x. The steps would be:

  1. Take the derivative of each term: 2x + 2y * dy/dx = 0.
  2. Move all terms with dy/dx to one side: 2y * dy/dx = -2x.
  3. Isolate dy/dx: dy/dx = -2x/2y = -x/y.

What are some real-world applications of implicit differentiation of many variables?

Implicit differentiation of many variables is used in various fields such as physics, economics, and engineering to analyze complex relationships between multiple variables. It is used to find rates of change, optimize functions, and solve differential equations. Some specific applications include studying the motion of objects in physics, maximizing profits in economics, and designing structures in engineering.

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