Derivative of the determinant of the right stretch tensor I have to evaluate the derivative
$$
\frac{\partial\det\mathcal{U}}{\partial F}
$$
where $\mathcal{U}=\sqrt{F^TF}$ and $F$ is a $m\times n$ real matrix. Any suggestion would be appreciated.
Thank you all, guys!! You helped me a lot.
 A: You can also approach the problem using differentials instead of the chain rule. It is easy to work with differentials, because algebraically they act like ordinary matrices.
Define a new matrix variable $W$, and its differential $$\eqalign{
W &= U^2 = F^TF = W^T \cr
dW &= 2\,\operatorname{sym}(F^T\,dF)\cr
}$$where $\operatorname{sym}(A)=\frac{1}{2}\,(A+A^T)$ is the symmetrization operation.
Now you want to find the gradient of
$$\eqalign{
g &= \det(U) = \sqrt{\det W} \cr
}$$
but it's more convenient work with its logarithm instead
$$\eqalign{
h &= \log(g) = \frac{1}{2}\,\log\det W\cr
dh &= d\log(g) = \frac{1}{2}\,d\operatorname{tr}\log W \cr
\frac{dg}{g} &= \frac{1}{2}\,W^{-1}:dW \cr
dg &= \frac{g}{2}\,W^{-1}:dW \cr
   &= g\,W^{-1}:\operatorname{sym}(F^T\,dF) \cr
   &= g\,\operatorname{sym}(W^{-1}):F^T\,dF \cr
   &= g\,W^{-1}:F^T\,dF \cr
   &= g\,FW^{-1}:dF \cr
\cr
}$$ where colon denotes the double-dot (aka Frobenius) product, which can be defined as $$A:B=\operatorname{tr}(A^TB)$$
So the gradient of interest is
$$\eqalign{
\frac{\partial g}{\partial F} &= gFW^{-1} \cr
  &= F(F^TF)^{-1}\det\sqrt{F^TF} \cr
}$$
A: Hint.
Name $f(F) = \det \sqrt{F^T F}$ You can write $f= f_3 \circ f_2 \circ f_1$ as the function composition of
$$\begin{array}{l|rcl}
f_1 : & \mathcal M_{m,n}(\mathbb R) & \longrightarrow & \mathcal S^+(\mathbb R^n) \subseteq \mathcal M_{n}(\mathbb R) \\
    & F & \longmapsto & F^T F \end{array}$$
$$\begin{array}{l|rcl}
f_2 : & \mathcal S^+(\mathbb R^n) & \longrightarrow & \mathcal S^+(\mathbb R^n) \\
    & M & \longmapsto & \sqrt{M} \end{array}$$
$$\begin{array}{l|rcl}
f_3 : & \mathcal M_{n} & \longrightarrow & \mathbb R \\
    & M & \longmapsto & \det M \end{array}$$
And finally
$$\begin{array}{l|rcl}
f : & \mathcal M_{m,n}(\mathbb R) & \longrightarrow & \mathbb R \\
    & F & \longmapsto & \det \sqrt{F^T F} \end{array}$$
You want to compute the Fréchet derivative $f^\prime$ of $f$. Applying the chain rule twice, you have:
$$f^\prime(F)=(f_3^\prime((f_2 \circ f_1)(F))) \circ f_2^\prime(f_1(F)) \circ f_1^\prime(F)$$
Now you have to find the Fréchet derivative of $f_1$, $f_2$ and $f_3$.
You have (Jacobi Formula):
$$f_3^\prime(F).H = \text{tr}(\text{adj}(F)H)$$ and (based on the derivative of a bilinear function between vector spaces)
$$f_1^\prime(F).H=F^T H + H^T F$$
The most difficult is to compute $f_2^\prime$. That can be done using Implicit function theorem, knowing that
$$(f_2(F))^2=F^T F$$ See Derivative (or differential) of symmetric square root of a matrix for more details on that last point.
