I am a bit confused on to how to prove differentiability in higher dimensions. My understanding is this so far:

1) If partial derivatives of a function exist and are continuous then it follows that the function is differentiable.

2) A function can be differentiable at a point without having its partial derivatives continuous at that point

3) A function is partially differentiable at a point a if and only if:

lim(x->a) (f(x)-f(a)-L(a)*(x-a)) / (x-a) = 0

Where L(a) is a linear function wich is the derivative of of f(a) in R1, the gradiant of f(a) in R2 and the Jacobian of f(a) in Rn. Is this correct?

So by considering all of this, a function can be differentiable at a point and discontinuous everywhere else.

  • 1
    $\begingroup$ differentiabilty in a point implies continuity in that same point. $\endgroup$
    – Thomas
    May 5 '16 at 17:39

In the definition

$$ \lim_{x \to a} \frac{\| f(x) - f(a) - L(x - a)\|}{\|x - a\|} = 0$$

It is clear that one must have $\lim_{x \to a} f(x) = f(a)$ since otherwise the numerator alone wouldn't even converge to 0 (as $x \to a$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.