Partial Derivative Chain Rule

  • #1
129
0

Homework Statement



Say that [tex]f(x)[/tex] is some function whose second derivative exists and say [tex]u(x, t)=f(x + ct)[/tex] for [tex]c > 0[/tex]. Determine

[tex]\frac{\partial u}{\partial x}[/tex]

In terms of [tex]f[/tex] and its derivatives.

Homework Equations



PD Chain rule.

The Attempt at a Solution



Say that x and y are both functions of f, ie. [tex]x = x(f)[/tex] and [tex]y = y(f)[/tex]. Then,

[tex]\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}u(x(f),y(f))\frac{dx}{d f}(x(f))[/tex]

[tex]\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}u(x,y)\frac{dx}{d f}(x(f))=\frac{\partial }{\partial x}f(x+ct)\frac{dx}{d f}(x+ct)[/tex]

[tex]\frac{\partial u}{\partial x}=f'(x+ct)[/tex]

Does that make any sense? It's mainly the notation that I don't understand, especially the "let x and y be functions of the same variable part."
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
No. That doesn't make any sense at all. You said f is a function of x+ct. That x would then be a function of f is crazy talk. What you've got there is just finding the partial derivative of f(x+ct) with respect to x where x and t are the independent variables.
 
  • #3
129
0
Thanks for the reply. So should I just disregard that extra function part and just write

[tex]\frac{\partial }{\partial x}u(x,t)=\frac{\partial }{\partial x}f(x+ct)=\frac{\partial }{\partial x}f(x+ct)\frac{d }{d x}(x+ct)=f'(x+ct)[/tex]

?
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
Thanks for the reply. So should I just disregard that extra function part and just write

[tex]\frac{\partial }{\partial x}u(x,t)=\frac{\partial }{\partial x}f(x+ct)=\frac{\partial }{\partial x}f(x+ct)\frac{d }{d x}(x+ct)=f'(x+ct)[/tex]

?
Yes, exactly.
 
  • #5
129
0
Okay. Thanks again :smile:
 

Related Threads on Partial Derivative Chain Rule

  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
784
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
846
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Top