Proving Differentialbility of h: R^n -> R^m | Step-by-Step Guide

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In summary, the individual trying to prove that a function is differentiable must use the coordinate projection function and the chain rule for one implication. If all components are differentiable, then the function itself is differentiable.
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Pearce_09
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Hello
Im trying to prove that a function h: R^n -> R^m is differentialbe if and only if each of the m components hi: R^n -> R is differentiable.
I know that i have to use the coordinate projection function and the chain rule for one implication, but I am having lots of trouble starting the problem off.
thanks
A.P.
 
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  • #3
If all components are differentiable then my guess is that you can use the triangle inequality to show that the function itself is dif'able.

For the only if part, you could assume that one of the components is not differentiable while the function is dif'able and hopefully a contradiction will be obvious.
 
  • #4
Exactly what is your definition of "differentiable"?
 
  • #5
Ok i used the triangle inequality to show that the function itself is differentiable; my work looks like
assume (h1, h2, ... , hm) is differentiable
therefore by the triangle inequality,
= ||(h1 ... hm)(x) - (h1...hm)(Xo) - D(h1...hm)(Xo)(x-Xo)||/ ||x-Xo||
and since h= (h1,h2, ..., hm)
then
||h(x) - h(Xo) - Dh(Xo)(x-Xo)|| / ||x-Xo||

now my proof was that the function is differentiable if and only if each component is differentiable.
and i assumed that one component was not differentable where the entire fn was differentiable. To hopefully find that it is a contradiction.
my work looks like: assume hj is not differentiable

since h= (hi,h2,..hj...,hm)
= ||(hi,...hj...hm)(x) - (hi,...hj,...,hm)(Xo) - D(hi,...hj,...,hm)(Xo)(x-Xo)||
/ ||x-Xo||

and i went on to say that this is
<= || hi(x) - hi(Xo) - Dhi(x-Xo) + ...+ hj(x) - hj(Xo) - Dhj(x-Xo) +...
but i assumed hj was not differentiable
therefor this is a contradiction, therefor if h is differentiable
then the,
sum (i=1 to m) hi must be differentiable


is this a good proof showing that all components are differentiable.. or am i doing somthing wrong?
 

FAQ: Proving Differentialbility of h: R^n -> R^m | Step-by-Step Guide

How do you define the differentialbility of a function h from R^n to R^m?

The differentialbility of a function h from R^n to R^m refers to the measure of how much the output of the function changes for a small change in the input. In other words, it is the ability of the function to be approximated by a linear transformation at a specific point.

What are the necessary conditions for a function to be differentialble?

In order for a function h from R^n to R^m to be differentialble, it must be continuous and have partial derivatives for all variables in the input. Additionally, the partial derivatives must be continuous at the point of interest.

How do you prove the differentialbility of a function h using the limit definition?

In order to prove the differentialbility of a function h using the limit definition, we need to show that the limit of the difference between the actual value of the function and its linear approximation, divided by the distance between the point and the point of interest, approaches 0 as the distance approaches 0.

Can a function be differentialble at a point but not continuous?

No, a function cannot be differentialble at a point where it is not continuous. In order for a function to be differentialble, it must also be continuous at that point.

What is the geometric interpretation of the differentialbility of a function?

The geometric interpretation of the differentialbility of a function is that at a specific point, the function can be approximated by a linear transformation, which is represented by the tangent plane to the graph of the function at that point.

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