# How to derive this complex differentiation?

In the finite-deformation theory, the elastic Cauchy-Green strain $$\mathbf{E}_e$$ is defined as

$$\mathbf{E}_e=\frac{1}{2}(\mathbf{F}_e^T \mathbf{F}_e-\mathbf{I})$$,

where the superscript $$T$$ denotes the transpose, $$\mathbf{F}_e$$ is the elastic component of the deformation gradient tensor $$\mathbf{F}$$ and $$\mathbf{I}$$ is the identity tensor.

With reference to the multiplicative decomposition of the deformation gradient $$\mathbf{F}$$, i.e. $$\mathbf{F}=\mathbf{F}_e\mathbf{F}_p$$, where $$\mathbf{F}_p$$ is the plastic part of the deformation gradient, $$\mathbf{F}_e$$ is calculated as

$$\mathbf{F}_e=\mathbf{F}\mathbf{F}_p^{-1}$$.

One can assume the formulation of $$\mathbf{F}_p$$ as following

$$\mathbf{F}_p=\exp((1-p_0)(p \log \mathbf{U}_I+(1-p)\log \mathbf{U}_J))$$,

where $$p_0$$ and $$p$$ are the plastic variables and $$\mathbf{U}_I$$ and $$\mathbf{U}_J$$ are the plastic stretch tensors, which are tensors possessing real numbers. Now, the question is how the differentiation of $$\mathbf{E}_e$$ with respect to $$p_0$$, i.e. $$\displaystyle \frac{\partial \mathbf{E}_e}{\partial p_0}$$, can be calculated.

For ease of typing, denote the operation $$\frac{\partial}{\partial p_0}$$ by $$d$$ \eqalign{ A &= p\log U_I + (1-p)\log U_J \cr F_p &= e^{(1-p_0)A} \cr dF_p &= -A F_p \cr\cr F_e &= FF_p^{-1} \cr dF_e &= F\,dF_p^{-1} \cr &= -FF_p^{-1}\,dF_p\,F_p^{-1} \cr &= FF_p^{-1}\,A F_p\,F_p^{-1} \cr &= F_eA \cr\cr E_e &= \tfrac{1}{2}(F_e^TF_e-I) \cr dE_e &= \tfrac{1}{2}(F_e^T\,dF_e+dF_e^T\,F_e) \cr &= \tfrac{1}{2}(F_e^TF_eA+A^TF_e^TF_e) \cr &= \tfrac{1}{2}((2E_e+I)A+A^T(2E_e+I)) \cr &= (E_eA + A^TE_e) + \tfrac{1}{2}(A+A^T) \cr }