Applying the chain rule using trees?

In summary, the conversation discusses the process of writing out a tree of dependencies and using it to find an expression for ## \partial z/\partial r ##. The person drew a tree and followed relevant paths, but only calculated two terms when there should be eight. The expert suggests using four terms instead, with three terms involving partial derivatives and one involving the partial of z with respect to y.
  • #1
Pizzerer
10
0

Homework Statement


Write out a tree (this will be a big tree) of dependencies and hence write down an expression for
∂z/∂r

Homework Equations


z=k(x, y)=xy2, x=(w1)(w2)+w3, y=w4;

w1=t, w2=t2, w3=2t+1, w4=sin(t);
t=r2+2s2

The Attempt at a Solution


This is the tree I drew and followed the relevant paths in order to try and write and expression but my expression is to short. I only calculate 2 terms when there is suppose to be 8:
http://www.picpaste.com/Untitled_Image-OZCiTMvb.jpg

This is wrong however. I need 16 leaves at the base. What am I doing wrong?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Pizzerer said:

Homework Statement


Write out a tree (this will be a big tree) of dependencies and hence write down an expression for
∂z/∂r

Homework Equations


z=k(x, y)=xy2, x=(w1)(w2)+w3, y=w4;

w1=t, w2=t2, w3=2t+1, w4=sin(t);
t=r2+2s2

The Attempt at a Solution


This is the tree I drew and followed the relevant paths in order to try and write and expression but my expression is to short. I only calculate 2 terms when there is suppose to be 8:
http://www.picpaste.com/Untitled_Image-OZCiTMvb.jpg

This is wrong however. I need 16 leaves at the base. What am I doing wrong?

I drew a diagram and got the same one you did.

As far as your partial ## \partial z/\partial r## is concerned, I get four terms to your two. Three of the four terms look like this:
$$ \frac{\partial z}{\partial x} \frac{\partial x}{\partial w_i} \frac{d w_i}{dt} \frac{\partial t}{\partial r}$$

For the fourth one, the first factor is the partial of z with respect to y.
 
Last edited by a moderator:
  • #3
attachment.php?attachmentid=58345&stc=1&d=1367266069.jpg
 

Attachments

  • Untitled_Image-OZCiTMvb.1367266012.jpg
    Untitled_Image-OZCiTMvb.1367266012.jpg
    15.7 KB · Views: 428

1. What is the chain rule in calculus?

The chain rule is a mathematical concept in calculus that allows us to find the derivative of a composite function, which is a function made up of two or more functions. It states that the derivative of a composite function is equal to the product of the derivatives of each individual function.

2. How is the chain rule applied using trees?

The chain rule can be applied using trees by representing the composite function as a tree diagram. Each branch of the tree represents one of the functions involved in the composite function, and the derivative of each function is written next to its corresponding branch. The final derivative is then calculated by multiplying the derivatives along the path from the input variable to the output variable.

3. What are the advantages of using trees to apply the chain rule?

Using trees to apply the chain rule allows for a visual representation of the composite function and its various components. This can help with understanding the relationship between the functions and their derivatives. It also allows for a systematic approach to calculating the derivative, making it less prone to mistakes.

4. Can the chain rule be applied to any composite function?

Yes, the chain rule can be applied to any composite function, as long as each individual function is differentiable. This means that the function must have a defined derivative at every point in its domain.

5. How can the chain rule be used in real-world applications?

The chain rule is used in many real-world applications, such as in physics, engineering, economics, and statistics. For example, it can be used to calculate the rate of change of a quantity in a system that is affected by multiple variables, or to optimize a process by finding the maximum or minimum value of a function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
909
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
976
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top