Derivative of an eigenvalue with respect to tensor itself

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I have a rank-2 tensor $\mathbf{C}=C_{\alpha\beta} \mathbf{A}^\alpha \otimes \mathbf{A}^\beta$ defined in a curvilinear coordinate system. I want to compute the derivative of an eigenvalue $\Lambda_i$ of the tensor w.r.t. to the tensor itself.

$\frac{\partial \Lambda_i}{\partial \mathbf{C}} = \frac{\partial \Lambda_i}{\partial C_{\alpha\beta}}\mathbf{A}_\alpha \otimes \mathbf{A}_\beta$

My approach is to use the relation $1 = \frac{\partial \Lambda_i}{\partial \Lambda_i} = \frac{\partial \Lambda_i}{\partial \mathbf{C}} : \frac{\partial \mathbf{C}}{\partial \Lambda_i}$. Then I found an expression for $\frac{\partial \mathbf{C}}{\partial \Lambda_i}$ and used $\mathbf{A} : \mathbf{A}^{-T} = \mathbf{A} \mathbf{A}^{-1} : \mathbf{I} = tr(\mathbf{I} ) = n $ to obtain $ \frac{\partial \Lambda_i}{\partial \mathbf{C}} = \frac{1}{n} (\frac{\partial \mathbf{C}}{\partial \Lambda_i})^{-T}$

Here, $n$ is the space dimension.

I am not sure whether this approach is correct, can anyone confirm this result?

Edit: For this to work, I assumed $\frac{\partial \mathbf{C}}{\partial \Lambda_i}$ invertible.