# Is the $n$-th order derivative the derivative of the $n-1$-th order one?

From Wikipedia:

The natural analog of second, third, and higher-order total derivatives is not a linear transformation, is not a function on the tangent bundle, and is not built by repeatedly taking the total derivative.

I think the $n$-th order (total) derivative of a mapping is the derivative of its $n-1$-th order derivative. In other words, derivative can be defined recursively. So I was wondering how to understand that "the natural analog of second, third, and higher-order total derivatives ... is not built by repeatedly taking the total derivative"?

Thanks and regards!

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I just thought the same thing about that Wikipedia article. As it is it looks plain wrong and Didier's answer below corroborates this. – Giuseppe Negro Mar 21 '12 at 0:51

As explained on the WP page, if $f:E\to F$ for finite dimensional vector spaces $E$ and $F$, then $f':E\to V$ where $V=L(E,F)$ is the space of linear maps from $E$ to $F$. Hence $f'':E\to W$ where $W=L(E,V)$ is the space of linear maps from $E$ to $V$, and so on.
In general, the vector spaces $V$, $W$ and their analogues for higher derivatives are all different, for example if $E$ and $F$ have dimensions $n$ and $m$, the dimension of $V$ is $nm$, the dimension of $W$ is $n^2m$, and so on, hence the objects $f'$, $f''$ and so on, are each in a different space. However, when $E=\mathbb R$, then for every finite dimensional vector space $U$ there is a natural identification of $L(E,U)=U^*$ with $U$ since every $\varphi$ in $L(\mathbb R,U)$ is $\varphi:x\mapsto xu$ for a given $u$ in $U$.
Using these identifications, when $E=\mathbb R$ and $F$ is finite dimensional, for each $n\geqslant1$, $f^{(n)}$ corresponds to a function from $\mathbb R$ to $L(\mathbb R,F)=F^*\simeq F$.