Derivative with Respect to Symmetric Tensor If you have a tensor $T$ with components $T_{ij}$, it seems clear that
$$
\frac{\partial T_{ij}}{\partial T_{k\ell}} = \delta^k_i \delta^{\ell}_j.
$$
However, if $T$ is symmetric, so $T_{ij} = T_{ji}$, there are two ways this derivative could be nonzero, so
$$
\frac{\partial T_{ij}}{\partial T_{k\ell}} = \delta^k_i \delta^{\ell}_j + \delta^\ell_i \delta^k_j.
$$
Does this mean the first expression is wrong? It seems that some information is lost if one naively uses the first expression with a symmetric matrix, because the result lacks the symmetry in $i$ and $j$ and in $k$ and $\ell$ that the derivative must have.
Additionally, should there be a factor of $\frac{1}{2}$? If one sets $i = j = k = \ell$, then the second expression yields 2. However, introducing this factor seems to then yield $\frac{1}{2}$ when $i = k$ and $j = \ell$ but $i \neq j$, when it should yield 1. What am I missing?
 A: I think this answers your question...
The $T_{ij}$ appear to be a basis for the dual of some vector space $V$. Let's assume that $V = \mathbb{C}^4 \otimes \mathbb{C}^4$, though this depends on the transformation rules (e.g. there may be duals in this formula). In general you can differentiate functions on $V$ in the direction of some vector $v \in V$, at a point $p \in V$. To differentiate $T_{ij}$ (a function) in the $T_{ij}$ direction means that you have identified $V^*$ with $V$ by some means. 
Moreover, there are two constructions of symmetric tensor -- first construction : as the subspace of $V$ cut out by the equations $T_{ij} = T_{ji}$, equivalently the space spanned by tensors of the form $e_i \otimes e_j + e_j \otimes e_i$ - lets call this $S(V)$. Second construction: As the quotient of $V$ by the space of tensors of the form $e_i \otimes e_j - e_j \otimes e_i$, lets call this $V/A(V)$ .... $A(V) \subset V$ is the space of these alternating tensors, $e_i \otimes e_j - e_j \otimes e_i$. (These spaces are isomorphic because there is a decomposition $V = S(V) \oplus A(V)$.) When you dualize subobjects turn into quotients, though we can again make identifications, it is important to be clear on this point. We are going to work with $S(V)$ throughout.
Anyway, this is all not helpful with a concrete computation, so let's follow some explicit tensors through this process. 
So let's suppose that we have an identification of $V$ with $V^*$, sending the basis elements $e_i \otimes e_j$ to the basis elements $e_i^* \otimes e_j^* = T_{ij}$. (Most likely this is what you do, since you call the $T_{ij}$ the components of the tensor, which is to say -- the coefficients when expanded in this standard basis $e_i \otimes e_j$.)
So, yes it is true that $\partial e_i^* \otimes e_j^* / \partial e_s \otimes e_t = \delta_{is} \delta_{jt}$.
Let's now work with $S(V)$. (Remember, $T_{ij}$ is a functional on $V$, so it restricts to be a functional on $S(V)$. This restriction is what identifies the dual of $S(V)$ as a quotient of $V^*$, where we mod out by the functionals that vanish on $S(V)$, aka $S(V)^{\perp}$. $T_{ij}$ and $T_{ji}$ are equal as functionals on $S(V)$, so they are equal in $V^* / S(V)^{\perp}$.)
Note that if we are identifying $S(V)$ with $S(V)^*$, then one option we have is to identify $T_{ij} = T_{ji} \in S(V)^*$ with $ (e_i \otimes e_j + e_j \otimes e_i)$. $T_{ij}(e_i \otimes e_j + e_j \otimes e_i) = 1$.
If we are working with a symmetric tensor in $V$, let's assume for concreteness that it is $v = e_s \otimes e_t + e_t \otimes e_s$. (This is identified with $T_{st}$ as in the previous paragraph.)  What if we differentiate the functional $T_{ij}$ in this direction? Recall that we said that $T_{ij} = T_{ji}$ on $S(V)$, so the answer should be the same differentiating $T_{ji}$ in the $v$ direction. Also note that it shouldn't matter if we differentiate along $v$ in $V$, so begin with we can compute $\nabla T_{ij}$, which is $e_i \otimes e_j$ (we are again implicitely using the identification of $V$ with $V^*$, since the differnetial of a function $f$ is a covector $df$, and eats vectors $w$ by differentiating $f$ in the direction of $w$). So $\partial T_{ij} / \partial v = \langle \nabla T_{ij} , v \rangle = \delta_{is} \delta_{jt} + \delta_{it} \delta_{js}$.
Okay, but you tried to write $\partial T_{ij} / \partial T_{kl}$. And you said that $T$ was a symmetric tensor. One way to interpret this is that you are thinking of $T$ as a function on the space of symmetric tensors. If you want to differentiate against $\partial  T_{kl}$ you have to turn it into a symmetric tensor -- or possibly just a tensor. As we have seen, there are two different choices for doing so -- that is, there is some ambiguity here. And it is the details of this choice that lead to your contradiction, since depeneding on the choices of your identification we get one of the two answers above.
So, suppose that by $\partial T_{kl}$ you mean $\partial (e_k \otimes e_l + e_l \otimes e_k)$. As discussed above (when we identified $S(V)$ with $S(V)^*$), $\partial T_{ij} / \partial T_{kl}$ is then equal to $\delta_{is} \delta_{jt} + \delta_{it} \delta_{js}$. On the other hand, if by $\partial T_{kl}$ you mean $\partial e_k \otimes e_l$ (as was the case when we discussed the first identification of $V$ with $V^*$), then $\partial T_{ij} / \partial T_{kl} = \delta_{is} \delta_{jt}$.
