Suppose ddz(f(z)) exists at a point z. Is ddz(f(z)) continuous at z?

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Linda Birchfield · Accepted Answer

Intuitively, you would maybe think so but the answer is no.As a counterexample, considerg(x):={x2sin⁡(1/x),x≠00x=0Then the derivative isg′(x)={2x2sin⁡(1/x)−cos⁡(1/x),x≠00x=0Note that limx→0g′(x) does not exist, even though g′(0)=0.

zesponderyd · Accepted Answer

If you are thinking to functions over the complex numbers, it is a celebrated result of Goursat that if the derivative exists over an open set, then the function is analytic over the open set; in particular its derivative is analytic as well, in particular differentiable and, therefore, continuous.   However, the derivative might not exist over an open set. An example is the function f(z)=|z|2=zz―, which is only differentiable at zero; indeed, the Wirtinger derivative is  ∂f∂z―=z  and a function can be differentiable (over the complex numbers) only where the Wirtinger derivative vanishes (it’s just the Cauchy-RIemann condition expressed in a different way). The derivative actually exists at zero, because  limz→0f(z)−f(0)z−0=limz→0z―=0  The derivative only exists at a point so it is continuous over its domain, but I guess it’s not the notion of continuity you were thinking to.

Suppose \frac{d}{dz}(f(z)) exists at a point z. Is \frac{d}{dz}(f(z)) continuous

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