How to understand this derivation - derivative of double dot product I am looking at a derivation from a continuum mechanics book and I am not sure I understand how the author goes from step 2 to step 3.  The author goes like this:


*

*$\frac{\partial tr[\bar{\textbf{C}}]}{\partial \textbf{C}}$

*$\frac{\partial [I_3^{-1/3}\textbf{C}:\textbf{I}]}{\partial \textbf{C}}$

*$I_3^{-1/3}\textbf{I}-\frac{1}{3}I_3^{-4/3}I_3\textbf{C}^{-1}(\textbf{C}:\textbf{I})$
I would think it would be easier to go from the second step to:
3'. $\frac{\partial [I_3^{-1/3}I_1]}{\partial \textbf{C}}$ 
Because $\textbf{C}:\textbf{I} = tr[\textbf{C}] = I_1$ 
Is this a valid move? If I do that I end up at the same final answer as the author, but I want to make sure it isn't blind luck.  If we follow the author's approach, it looks like one needs to apply the product rule to the three terms that are each a function of $\textbf{C}$ but I am not sure how this works out.  For example, applying the product rule to the second step gives:
2'. $\frac{\partial I_3^{-1/3}}{\partial \textbf{C}}\textbf{C}:\textbf{I}+I_3^{-1/3}\frac{\partial \textbf{C}}{\partial \textbf{C}}:\textbf{I}+I_3^{-1/3}\textbf{C}:\frac{\partial \textbf{I}}{\partial \textbf{C}}$
The third term will be zero because the identity matrix is independent of $\textbf{C}$.  But does the author's step 3 imply that $\frac{\partial \textbf{C}}{\partial \textbf{C}}:\textbf{I} = \textbf{I}$?  I thought that $\frac{\partial \textbf{C}}{\partial \textbf{C}}$ was the 4th order identity?  
The final expression is:


*$I_3^{-1/3}(\textbf{I}-\frac{1}{3}I_1\textbf{C}^{-1})$
So in summary, my questions are:
Is my method (3') valid? 
Do I understand correctly what the author's work implies about the inner product of a 4th order identity with a 2nd order identity being a second order identity?
Thanks
 A: Replacing $\operatorname{tr}\boldsymbol C$ by $\boldsymbol C:\boldsymbol I$ is perfectly fine and it is not just a fluke that you end up with the correct answer.
Concerning the expression
$
\frac{\partial\boldsymbol C}{\partial\boldsymbol C}:\boldsymbol I, 
$
maybe it becomes clearer if we use a basis representation. I assume $\boldsymbol C$ is a symmetric tensor. Then 
$$
\frac{\partial\boldsymbol C}{\partial\boldsymbol C}=\stackrel{<4>}{\boldsymbol I_\mathrm{sym}}
$$
which is the fourth order symmetric unit tensor. It has the following basis representation in an orthonormal basis $\{\boldsymbol e_i\}$:
$$
\stackrel{<4>}{\boldsymbol I_\mathrm{sym}}=\frac12\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)
\boldsymbol e_i\otimes \boldsymbol e_j\otimes \boldsymbol e_k \otimes \boldsymbol e_l\,.
$$
With $\delta_{ij}$ being the usual Kronecker-$\delta$.
The second order unit tensor basis representation reads
$$
\boldsymbol I =\delta_{mn}\boldsymbol e_m\otimes \boldsymbol e_n\,.
$$
The double contraction (from the right side) then is
\begin{align}
\stackrel{<4>}{\boldsymbol I_\mathrm{sym}}:\boldsymbol I &=
\frac12\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)\delta_{mn}
\left[\boldsymbol e_i\otimes \boldsymbol e_j\otimes \boldsymbol e_k \otimes \boldsymbol e_l\right]:\left[\boldsymbol e_m\otimes \boldsymbol e_n\right]\\
&=\frac12\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)\delta_{mn}\left[\boldsymbol e_i\otimes \boldsymbol e_j\right]\left(\boldsymbol e_k \cdot \boldsymbol e_m\right)\left(\boldsymbol e_l\cdot \boldsymbol e_n\right)\\
&=\frac12\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)\delta_{mn}\delta_{km}\delta_{ln}\left[\boldsymbol e_i\otimes \boldsymbol e_j\right]\\
&=\frac12\left(\delta_{ij}+\delta_{ji}\right)\left[\boldsymbol e_i\otimes \boldsymbol e_j\right]\\
&=\delta_{ij}\left[\boldsymbol e_i\otimes \boldsymbol e_j\right]\\&=\boldsymbol I
\end{align}
An easier way to think of it is that every fourth order tensor induces a linear mapping from the space of second order tensors to the space of second order tensors.
The symmetric fourth order unit tensor maps every second order tensor onto its symmetric part (and thus any symmetric second order tensor onto itself).
A: The Wikipedia article on tensor derivatives contains a simple proof of assertion 3, showing that $\frac{\partial C}{\partial C}$ is indeed the fourth order identity tensor and the contraction of the fourth and second order identity tensors produces the second order identity tensor.
