Proof of $\frac{\partial {det(\mathbf X)}}{\partial \mathbf X}=det(\mathbf X)(\mathbf X^{-1})^*$ let X be a square, complex, and Hermitian matrix. Show that 
$\frac{\partial {det(\mathbf X)}}{\partial \mathbf X}=det(\mathbf X)(\mathbf X^{-1})^*$ and $\frac{\partial {det(\mathbf X)}}{\partial \mathbf X^*}=det(\mathbf X)(\mathbf X^{-1})$
Can anyone teach me how to prove this?
 A: Assuming that $X$ is invertible, we have
$$\det(X + \epsilon) = \det(X(1 + X^{-1} \epsilon)) = \det(X) \det(1 + X^{-1} \epsilon) = \det(X) (1 + \text{tr}(X^{-1} \epsilon)).$$ 
Here $\epsilon$ is an infinitesimal matrix. The derivative of $\det X$, as a function from $n \times n$ matrices to scalars, is a linear functional from $n \times n$ matrices to scalars, and these can be identified with $n \times n$ matrices using the Hilbert-Schmidt inner product $\text{tr}(X^{\ast} Y)$; that is, the linear functional $Y \mapsto \text{tr}(X^{\ast} Y)$ gets identified with $X$. With respect to this identification, the above computation shows that the derivative of $\det(X)$ at $X$ is the linear functional
$$\epsilon \mapsto \det(X) \text{tr}(X^{-1} \epsilon)$$
and hence it gets identified with the matrix $\det(X) (X^{-1})^{\ast}$ as desired. There is no need to assume that $X$ is Hermitian.  
The same trick of using an infinitesimal matrix can be used to differentiate basically any matrix function. It can be made fully rigorous in several ways, and I have no idea why nobody seems to teach it. 
