Solving the Problem: Deriving ∂2φ/∂x2

  • Thread starter Thread starter whatisreality
  • Start date Start date
  • Tags Tags
    deriving
Click For Summary

Homework Help Overview

The problem involves deriving the second partial derivative of a function φ with respect to x, where φ is a function of r, and r is defined as the square root of the sum of the squares of x, y, and z. The goal is to express this second derivative in a specific form involving first and second derivatives of φ with respect to r.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the chain rule and product rule in finding the derivatives. There are attempts to express ∂φ/∂x using ∂r/∂x and concerns about accounting for all dependencies. Some participants express confusion about missing terms and the correct application of derivatives.

Discussion Status

Multiple interpretations of the derivative calculations are being explored, with participants questioning their own reasoning and the assumptions made about the variables involved. Some guidance is provided regarding the treatment of r as a function of x, y, and z, which has led to clarifications in the calculations.

Contextual Notes

Participants note the complexity of the problem due to the interdependence of the variables and the necessity of careful application of derivative rules. There is an acknowledgment of the specific form that the final expression must take, which adds to the challenge of the problem.

whatisreality
Messages
286
Reaction score
1

Homework Statement


Let φ=φ(r) and r=√x2+y2+z2. Find ∂2φ/∂x2.
Show that it can be written as (1/r + x2/r3)∂φ/∂r + x2/r22φ/∂r2.

Homework Equations


Use the identity ∂r/∂x = x/r.

The Attempt at a Solution


I think I know ∂φ/∂x. Using the chain rule, it's ∂r/∂x ∂φ/∂r. That gives x/r ∂φ/∂r. If that's wrong it might be because I know you have to take account of all dependences, but I don't actually know how to.
So assuming that's ok, I then need to use the product rule, and that gave me
1/r ∂φ/∂r + ∂φ/∂x ∂φ/∂r ∂r/∂x.
Which I know is wrong, because it's a show that question!
 
Physics news on Phys.org
whatisreality said:

Homework Statement


Let φ=φ(r) and r=√x2+y2+z2. Find ∂2φ/∂x2.
Show that it can be written as (1/r + x2/r3)∂φ/∂r + x2/r22φ/∂r2.

Homework Equations


Use the identity ∂r/∂x = x/r.

The Attempt at a Solution


I think I know ∂φ/∂x. Using the chain rule, it's ∂r/∂x ∂φ/∂r. That gives x/r ∂φ/∂r.
I would write this as dφ/dr ∂r/∂x or φ'(r)∂r/∂x = dφ/dr (x/r). φ is a function of r alone, so the derivative for this function is the regular derivative instead of the partial derivative. Now use the product rule to get ∂2φ/∂x2.
whatisreality said:
If that's wrong it might be because I know you have to take account of all dependences, but I don't actually know how to.
So assuming that's ok, I then need to use the product rule, and that gave me
1/r ∂φ/∂r + ∂φ/∂x ∂φ/∂r ∂r/∂x.
Which I know is wrong, because it's a show that question!
 
Mark44 said:
I would write this as dφ/dr ∂r/∂x or φ'(r)∂r/∂x = dφ/dr (x/r). φ is a function of r alone, so the derivative for this function is the regular derivative instead of the partial derivative. Now use the product rule to get ∂2φ/∂x2.
I'm doing something wrong. So I got ∂2φ/∂x2 = 1/r ∂φ/∂r + ∂/dr (dφ/dr) ∂r/∂x x/r. This doesn't give me the answer, it gives
2φ/∂x2 = 1/r dφ/dr + x2/r22φ/∂r2, but I really can't see what's wrong!
 
I'm missing a whole term: x2/r3 dφ/dr.
 
whatisreality said:
I'm doing something wrong. So I got ∂2φ/∂x2 = 1/r ∂φ/∂r + ∂/dr (dφ/dr) ∂r/∂x x/r. This doesn't give me the answer, it gives
2φ/∂x2 = 1/r dφ/dr + x2/r22φ/∂r2, but I really can't see what's wrong!
You don't show the work leading up to this, but I'm guessing that you did this: ##\frac{\partial}{\partial x} \frac{x}{r} = \frac{1}{r}##. If so, that's wrong, since you would be treating r as a constant. In fact, r is a function of x (and y and z).
 
Ohhh... That's exactly what I did. I always forget that!
 
Should be y2+z2/r3 then.
Yep, that gives the right answer! Thank you, I was getting really confused.
 

Similar threads

Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K