# When does a gradient vector of a function not exist?

I have a question that gives me a 3d function and asks me to calculate the gradient vector of it. This part I understand. It then asks me to indicate the points at which it does not exist. When does a gradient vector not exist? Is it when it equals to zero? Or does it mean when the function DNE.

Thanks.

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Well the gradient is defined as the vector of partial derivatives so that it will exist if and only if all the partials exist. A zero gradient is still a gradient (it's just the zero vector) and we sometimes say that the gradient vanishes in this case (note that vanish and does not exist are different things). A function can still be well defined at a point without the gradient existing, in fact a function can still be continuous without the gradient existing.

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One might define $\nabla f(x)$ to be the unique vector $g$ such that $f(x +h) =f(x) +g^T h +o(h)$ as $|h| \to 0$ (assuming such a $g$ exists). – littleO Nov 6 '12 at 22:03
Thanks! From what I understand tho, if a partial derivative of the gradient does not exist, it means that it is defined as a critical point. – Yamato C Nov 6 '12 at 22:07
@littleO That definition holds only for differentiable functions in which case it just reduces to the normal definition. – EuYu Nov 6 '12 at 22:11
Is this critical point the point where the gradient does not exist? I don't believe so since, from what I have learned, this point is either a local/absolute maximum, minimum or saddle point. – Yamato C Nov 6 '12 at 22:14
A critical point is typically defined as a point in which the gradient vanishes. It'll be a local max/min or saddle depending on the definiteness of the Hessian. – EuYu Nov 6 '12 at 22:17