What does it mean for a partial derivative to be continuous at a point?

  • Thread starter Thread starter Jamin2112
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the continuity of partial derivatives and their implications for differentiability at a point. Specifically, it references Theorem I, which states that if the partial derivative ∂f/∂x exists and is continuous at (a,b), while ∂f/∂y is defined at (a,b), then the function f is differentiable at that point. The conversation delves into the mathematical expressions used to illustrate this theorem, particularly the application of the mean value theorem and the behavior of the terms Ω1 and Ω2 as they approach zero.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with the concept of differentiability
  • Knowledge of the mean value theorem
  • Basic calculus, particularly limits and continuity
NEXT STEPS
  • Study the implications of the Mean Value Theorem in multivariable calculus
  • Explore the definition and properties of differentiability in higher dimensions
  • Investigate the relationship between continuity and differentiability of functions
  • Learn about the behavior of error terms in Taylor expansions for multivariable functions
USEFUL FOR

Students and educators in calculus, mathematicians focusing on multivariable analysis, and anyone seeking to deepen their understanding of continuity and differentiability in the context of partial derivatives.

Jamin2112
Messages
973
Reaction score
12
I'm trying to the follow the work shown in the textbook, but getting a little confused.

Theorem I. Suppose the function f(x,y) is defined in some neighborhood of the of the point (a,b). Suppose one of the partial derivatives, say, ∂f/∂x, exists at each point of the neighborhood and is continuous at (a,b), while the other partial derivative is defined at least at the point (a,b). Then f is differentiable at (a,b).

The first thing the book does is write down

f(a+h, b+k) - f(a,b) = f(a+h, b+k) - f(a, b+k) + f(a, b+k) - f(a,b).

Okay. So far, so good.

Now, the part I put in red looks like the [f(a,b+k)-f(a,b)]/k, the partial derivative of f with respect to y at (a,b) when k approaches zero. But since k is just k, we can write this as

f(a+h, b+k) - f(a, b+k) = k ∂f/∂y|(a,b) + Ω1k, where Ω1 approaches zero as k approaches 0.

The part I put in blue, I suppose, could similarly be rewritten as h ∂f/∂x|(a,b+k) + hΩ2. Instead the book uses the mean value theorem as such:

f(a+h, b+k) - f(a, b+k) = h ∂f/∂x|(a+øh, b+k), where 0<ø<1.

Because ∂f/∂x is assumed to be continuous at (a,b), we can write

∂f/∂x|(a+øh, b+k) = ∂f/∂x|(a,b) + Ω2, where Ω2 --> 0 as k --> 0 and h --> 0.

The part in green is what I do not understand. Please explain.
 
Physics news on Phys.org
That's just what it means for the partial derivative to be continuous at (a,b). Since ∂f/∂x|(a+øh, b+k) -> ∂f/∂x|(a,b) as h -> 0 and k -> 0, we can for example define Ω2 to be the difference ∂f/∂x|(a+øh, b+k) - ∂f/∂x|(a,b) and see that Ω2 -> 0 when h -> 0 and k -> 0.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
1K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K