Differential of the deviatoric tensor

Question

Given the relation: $$\pmb{S} = \pmb{T} + p\pmb{I} \\ p = -\frac{1}{3}tr\mathbf{T}$$ $\pmb{T}$ is the Cauchy stress tensor, $\pmb{S}$ is the deviatoric stress tensor, $\pmb{I}$ is the identity tensor and $p$ is the mean stress. How to derive the differential of the deviatoric stress tensor: $$\frac{\partial \mathbf{S}}{\partial p}$$Is the answer zero tensor? (If the question is not clear enough, please tell me...)

Answer 1

I think that the answer is a zero tensor, because of the following reasons.
Tensor space is also a vector space.
The inner product between an isotropic tensor and
a deviatoric tensor or a trace-less symmetric tensor is zero, i.e. $\boldsymbol I:\boldsymbol S=tr(\boldsymbol S)=0$, and the two tensors are orthogonal.
Therefore, the isotropic tensor field and the traceless symmetric tensor field are independent.