I have been trying to find an analytical expression for the following:
$\frac{\partial {X^{+}}}{\partial {X}}$
In my case, $X$ has a constant rank.
I've found the formula for differentiating a pseudoinverse in Goulob's paper (equation 4.12),:
$$
\frac{\mathrm d}{\mathrm d x} A^+(x) =
-A^+ \left( \frac{\mathrm d}{\mathrm d x} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d x} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d x} A^T \right) A{^+}^T A^+
$$
but I can't see how to input the original matrix:. I've tried
$$
\frac{\mathrm d}{\mathrm d x} A^+(x) =
-A^+ \left( \frac{\mathrm d}{\mathrm d x} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d x} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d x} A^T \right) A{^+}^T A^+
$$$$
\frac{\mathrm d}{\mathrm d A} A^+(A) =
-A^+ \left( \frac{\mathrm d}{\mathrm d A} A \right) A^+
+A^+ A{^+}^T \left( \frac{\mathrm d}{\mathrm d A} A^T \right) (1-A A^+)
+ (1-A^+ A) \left( \frac{\mathrm d}{\mathrm d A} A^T \right) A{^+}^T A^+
$$
but it doesn't seem to give the right answers. What should $\frac{\mathrm d A^T}{\mathrm d A}$ and $\frac{\mathrm d A}{\mathrm d A}$ evaluate to?
