# Continuous differentiability of vector fields

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My attempt:

a) F(x,y) = (f(x,y), g(x,y)) is differentiable at V = (a,b) if F(v + h) - F(v) = Jacobian(F) h + R(h) with the limit R(h)/norm(h) -> 0 as h -> 0

OR equivalently all partial derivates of F's components exist and are continuous in a neighbourhood of V

and F is continuously differentiable if all partial derivatives of each component exist and are continuous at V.

At (x,y) = (0,0), F isn't differentiable since approaching this point along the line y = x would yield the limit( (2h^2 / 2h^2) /h ) as h ->0 which is definitely not 0. So F isn't differentiable at (0,0)

For the rest of the domain (except this xy = 1 line and (0,0) ), F is continuously differentiable since the components of F are ratios of continously differentiable functions.

So the only place a problem could arise is this xy = 1 line:

xy - 1 < 0: the second component is 1-xy ,

d/dx(1-xy) = -y

xy -1 > 0: the second component is xy - 1

d/dx(xy-1) = y

so the partial derivatives of the 2nd component aren't continuous across the line.

I'm at a loss as to how I should show whether or not the function is differentiable across the line? It seems like it isn't since the partial derivates won't agree in any neighbourhood of a point in the line.

So would the answer to the question be no, because everywhere F isn't is differentiable it's continuously differentiable and everywhere it may not be continuously differentiable it's not even differentiable?

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#2

(Original post by

My attempt:

a) F(x,y) = (f(x,y), g(x,y)) is differentiable at V = (a,b) if F(v + h) - F(v) = Jacobian(F) h + R(h) with the limit R(h)/norm(h) -> 0 as h -> 0

OR equivalently all partial derivates of F's components exist and are continuous in a neighbourhood of V

and F is continuously differentiable if all partial derivatives of each component exist and are continuous at V.

**Ryanzmw**)My attempt:

a) F(x,y) = (f(x,y), g(x,y)) is differentiable at V = (a,b) if F(v + h) - F(v) = Jacobian(F) h + R(h) with the limit R(h)/norm(h) -> 0 as h -> 0

OR equivalently all partial derivates of F's components exist and are continuous in a neighbourhood of V

and F is continuously differentiable if all partial derivatives of each component exist and are continuous at V.

So would the answer to the question be no, because everywhere F isn't is differentiable it's continuously differentiable and everywhere it may not be continuously differentiable it's not even differentiable?

(The standard example of an F that's differentiable but not continuously so is to use the 1D function g(x) = x^2 sin(1/x) (with g(0) = 0) that's diff but not cts diff at x = 0, and then define F(x, y) = g(x) + g(y)).

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(Original post by

I don't agree with the definition I've highlighted in red (I can't see how it's different from the statement I've made green).

I agree that there's no point where F it differentiable without being continuously differentiable.

(The standard example of an F that's differentiable but not continuously so is to use the 1D function g(x) = x^2 sin(1/x) (with g(0) = 0) that's diff but not cts diff at x = 0, and then define F(x, y) = g(x) + g(y)).

**DFranklin**)I don't agree with the definition I've highlighted in red (I can't see how it's different from the statement I've made green).

I agree that there's no point where F it differentiable without being continuously differentiable.

(The standard example of an F that's differentiable but not continuously so is to use the 1D function g(x) = x^2 sin(1/x) (with g(0) = 0) that's diff but not cts diff at x = 0, and then define F(x, y) = g(x) + g(y)).

all continuous partial derivatives in a neighbourhood of a <=> Differentiable at a

Is there no alternative definition of differentiable in terms of partial derivatives? Or am I forced to give the first definition?

It's certainly possible I was mistaken in writing this down as a theorem though...

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#4

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I agree that they do seem to be the same thing... I was just rattling off a theorem given that:

all continuous partial derivatives in a neighbourhood of a <=> Differentiable at a

**Ryanzmw**)I agree that they do seem to be the same thing... I was just rattling off a theorem given that:

all continuous partial derivatives in a neighbourhood of a <=> Differentiable at a

Conversely, by considering at the origin you can see the partial derivatives are trivially zero, but by considering the behaviour as you approach the origin on the line y = x you can see that the directional derivative in this direction is 1/2. So this function can't be differentiable.

Basically:

"partial derivatives exist and are cts" is a stronger condition than

"differentiable", which is a stronger condition than

"partial derivatives exist".

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(Original post by

This theorem is not correct - I just gave you an example of a function differentiable at the origin but with discontinuous partial derivatives there.

Conversely, by considering at the origin you can see the partial derivatives are trivially zero, but by considering the behaviour as you approach the origin on the line y = x you can see that the directional derivative in this direction is 1/2. So this function can't be differentiable.

Basically:

"partial derivatives exist and are cts" is a stronger condition than

"differentiable", which is a stronger condition than

"partial derivatives exist".

**DFranklin**)This theorem is not correct - I just gave you an example of a function differentiable at the origin but with discontinuous partial derivatives there.

Conversely, by considering at the origin you can see the partial derivatives are trivially zero, but by considering the behaviour as you approach the origin on the line y = x you can see that the directional derivative in this direction is 1/2. So this function can't be differentiable.

Basically:

"partial derivatives exist and are cts" is a stronger condition than

"differentiable", which is a stronger condition than

"partial derivatives exist".

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#6

(Original post by

Thank you very much!

**Ryanzmw**)Thank you very much!

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(Original post by

If you can get access to it, the book "Counterexamples in Analysis" is *really* useful for these kinds of questions.

**DFranklin**)If you can get access to it, the book "Counterexamples in Analysis" is *really* useful for these kinds of questions.

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