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 [Golub's paper](http://i.stanford.edu/pub/cstr/reports/cs/tr/72/261/CS-TR-72-261.pdf) (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.



Derivative of pseudoinverse with respect to original matrix

> 
> but I can't see how to input the original matrix.
> 
> 
> 


I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in [this answer](https://math.stackexchange.com/questions/1689434/derivative-of-the-frobenius-norm-of-a-pseudoinverse-matrix))
$$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$


Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:


$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$


