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The general definition of a differentiable mapping is,

Let U be an open set in Rn, and let ‘a’ be in U and f:Rp. Then f is a differentiable mapping at ‘a’ if there exists a Df(a) in Hom(Rn, Rp) such that;

lim┬(x→a)⁡〖(||f(x)-f(a)-Df(a)(x-a) ||)/(||x-a||)〗=0

Why is U necessarily an open set?

Siddhu
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1 Answers1

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We need it to move "freely" and not to get to boundary of the $U$ where we cannot move there in the way we want. Another thing is that you need to have a nice and handy form of the function's domain, so you can choose between $U$ being open or closed in $\mathbb R^n$. Now if $U$ is closed then sometimes we can have one degree of freedom because of the boundary of $U$. So $f$ would map $U$ to $\mathbb R^{n-1}$ sometimes. To get pass this, we choose $U$ to be open ,so to the image of every element via $f$ would be an $n$ variable vector.

Haha
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