Let $f:\mathbb{R}\to \mathbb{R}$ be a function, then looking $f$ as a function between manifolds, $df:T\mathbb{R}=\mathbb{R}^2\to \mathbb{R}^2$ and $d^2f:TT\mathbb{R}=\mathbb{R}^4\to \mathbb{R}^4$ are given by


$$df(x,a)=(f(x),\frac{df}{dx}a)$$
$$d^2f((x,a),(b,c))= ((f(x),\frac{df}{dx}a),(\frac{df}{dx}b,\frac{d^2f}{dx^2} ab+ \frac{df}{dx} c))$$


So the second derivative of $f$, as you know it from calculus, is embedded in the differetial geometric definition of $d^2f$ in a wierd way. There is a similar story in higher orders.


Now suppose $f:M\to N$, $f(x)=y$, is a smooth map between two manifolds. As before, for each k, we have $d^kf: TTT\cdots TN \to TTT\cdots TM$. 


Question: For $v\_1,\cdots,v\_k \in T\_xM$, is there an object $\frac{d^kf}{dv\_1\cdots dv\_k}$ similar to the way we define higher derivatives of a function on $\mathbb{R}$? how is it defined? and where does it belong to? (Is $\frac{df}{dv\_1\cdots dv\_k}\in T\_yN$ ?) 


Apologies in advance if the question is obvious or absurd.



An elementary but confusing question in differential geomerty
It's not very clear to me which properties you expect from that object, but interpreting the question literally the answer is no: there is no natural way to associate to some vectors $v\_1,v\_2,\ldots,v\_k\in T\_x M$ and a function $f\in C^\infty(M)$ an object corresponding "only" to the partial derivative $\frac{\partial ^kf}{\partial v\_1\cdots\partial v\_k}$ $k\geq 2$. As Otis remarks there are jets, but these encode all lower order derivatives too, and there is no natural way to disentangle them. The reason is that all functions with $df(x)\neq 0$ look locally the same, so they would all have "second derivative" zero. This is what Will says in his remark.


On the other hand, if the function $f$ vanishes to order $k-1$ in $x$ (in terms of jets: makes $k-1$-contact with a constant function), then the partial derivative depends only on the vectors $v\_1,v\_2,\ldots,v\_k\in T\_x M$ and is an element of $T\_{f(x)}Y$.


If you assume a metric on $M$ you should also be able to save the situation, maybe extending the vectors to geoedesics and then by parallel transport to vector fields... but I'm not sure about that.


**Edit in response to the comment**: a high level answer is this: the fibers of the projection from the space of jets of order k to jets of order k-1 are naturally affine spaces over the vector space $S^k(T^\*\_x M)\otimes T\_y N$. So if two maps have contact of order k-1, then their jets of order k differ by a k-symmetric map $T\_x M\times \cdots T\_x M \to T\_y N$, which in your case is the map you expect in local coordinates $v\_1,\ldots ,v\_k\to \frac{\partial ^kf}{\partial v\_1\cdots\partial v\_k}$. There are different proofs for this affine structure in the literature on jets (personally I find non of them particularly enlightening).


In the case that $N=\mathbb{R}$ it is not so hard to show directly, that if you extend your vectors to vector fields $X\_1,\ldots,X\_2$ then the map $v\_1,\ldots ,v\_k\mapsto X\_k(X\_{k-1}(\cdots (X\_1(f))\cdots)(x)$ is independent of the choice of extensions when $f$ vanishes to order k-1 in $x$.


