Derivative of coefficients of characteristic polynomial Let $ X(t) $ be a square matrix of dimension $ n $. The Jacobi formula expresses the derivative of the determinant of $X(t)$ in terms of the derivative of the matrix itself. Is there an analogous formula for the other coefficients of the characteristic polynomial?
I am actually interested in the $ n = 3 $ case, so the only coefficient I do really care about is the one of $ t $.
Thank you very much!
 A: The Newton Identities can be used to write the characteristic polynomial of $X\in{\mathbb C}^{3\times 3}$ in terms of the traces of the powers of $X$
$$\eqalign{
p(X) &= X^3 +c_2X^2 +c_1X +c_0I \\
&= X^3 - X^2\operatorname{tr}(X) + X\left(\frac{\operatorname{tr}^2(X)-\operatorname{tr}(X^2)}{2!}\right) - I\det(X) \\
}$$
Calculate the differentials of the trace functions
$$\eqalign{
\operatorname{tr}(X) &= I:X \quad&\implies d\operatorname{tr}(X) = I:dX \\
& &\implies d\operatorname{tr}^2(X) = 2\operatorname{tr}(X)\,d\operatorname{tr}(X)
 = 2\operatorname{tr}(X)I:dX\\
\operatorname{tr}(X^2) &= I:X^2 \quad&\implies d\operatorname{tr}(X^2) = I:(X\,dX+dX\,X) = 2X^T:dX \\
}$$
where the colon is a convenient product notation for the trace, i.e.
$\;A:B = \operatorname{tr}(A^TB)$
Then calculate the differential, derivative (wrt $t$) and gradient (wrt $X$) of the $c_1$ coefficient 
$$\eqalign{
dc_1 &= \tfrac{1}{2}(2\operatorname{tr}(X)I-2X^T) : dX \\
\dot{c}_1 &= \big(\operatorname{tr}(X)I-X^T\big) : \dot{X} \\
\frac{\partial c_1}{\partial X} &= \operatorname{tr}(X)I-X^T \\
}$$
As a bonus, do the same for the $c_2$ coefficient 
$$\eqalign{
dc_2 &= -I:dX \\
\dot{c}_2 &= -I:\dot{X} \\
\frac{\partial c_2}{\partial X} &= -I \\
}$$ 
Even the determinant (the $c_0$ coefficient) can be written in terms of  traces 
$$\det(X) = 
\frac{\operatorname{tr}^3(X)
-3\operatorname{tr}(X)\operatorname{tr}(X^2)
+2\operatorname{tr}(X^3)}{3!}
$$
A: The 1st order coefficient in the (cubic) characteristic polynomial is given by $f : \mathbb R^{3 \times 3} \to \mathbb R$, defined by
$$f ({\rm X}) := \frac12 \mbox{tr}^2 \left( {\rm X} \right) - \frac12 \mbox{tr} \left( {\rm X}^2 \right)$$
as Greg mentioned. Hence,
$$\begin{aligned} f ({\rm X} + h {\rm V}) = \cdots &= f ({\rm X}) + h \left( \mbox{tr} \left( {\rm X} \right) \mbox{tr} \left( {\rm V} \right) - \mbox{tr} \left( {\rm V} {\rm X} \right) \right) + O (h^2)\\ &= f ({\rm X}) + h \left\langle \color{blue}{\mbox{tr} \left( {\rm X} \right) {\rm I}_3 - {\rm X}^\top}, {\rm V} \right\rangle + O (h^2)\end{aligned}$$
where the Frobenius inner product was used. Thus, the gradient of $f$ with respect to $\rm X$ is
$$\boxed{\nabla f ({\rm X}) = \mbox{tr} \left( {\rm X} \right) {\rm I}_3 - {\rm X}^\top}$$
which is the same Greg obtained.

matrix-calculus scalar-fields gradient
