A question about an identity in deriving Born-Infeld action I have a question in David Tong's Example Sheet 4
Problem 5b, how to verify the last equation (*) on p.2? (There is a solution for example sheet 3, but seems to be no solution for example sheet 4.)

Problem 5b:
Show that the equations of motion                         arising from the          Born-Infeld action are equivalent to the beta function condition for the open string,
  $$\beta_\sigma\left(F\right)=\left( \frac{1}    {1 -F^2} \right)^{\mu \rho}\partial_\mu     F_{\rho\sigma    }=0      $$
     Note: To do this, it will prove very useful if you can ﬁrst show the following results:
  $$∂_μ\left[\operatorname{tr}      \ln(1 − F^2)\right]           = −4 ∂_\rho    F_{μ\sigma}\left(\frac{F}{1-F^2}\right)^{\sigma \rho   }             
$$        
which requires use of the Bianchi identity for $F_{\mu \nu   }$    and       
$$ \tag{*}
\begin{align}
\partial _\mu      \left( \frac{ F}{1-F^2} \right)^{\mu\nu}
&=
    \left( \frac{ F}{1-F^2} \right)^{\mu\rho}   \partial_\mu F_{\rho\sigma}    \left( \frac{ F}{1-F^2} \right)^{\sigma    \nu} 
\\ &\qquad+    \left( \frac{1}    {1 -F^2} \right)^{\mu \rho} \partial_\mu F_{\rho\sigma}\left( \frac{ 1        }{1-F^2} \right)^{\sigma    \nu}             \end{align}$$       

In addition, as given in question 5a        

$$F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} $$

My attempt to prove the problem:
LHS 
$$\partial_{\mu} \left( \frac{ F}{1-F^2} \right)^{\mu\nu} = \partial_{\mu} \left[ F^{\mu}_{\alpha} \left( \frac{1}{1-F^2} \right)^{\alpha \nu} \right] = \left( \partial_{\mu}  F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + F^{\mu}_{\alpha} \partial_{\mu} \left( \frac{1}{1 -F^2}  \right)^{\alpha \nu} \tag{1} $$
Using the formula in matrix cookbook for the derivative of inverse matrix, Eq. (53) 
Eq. (1) becomes
$$\left( \partial_{\mu}  F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2 F^{\mu}_{\alpha} \left[ \frac{1}{1 -F^2} \left( \partial_{\mu} F \right) F \frac{1}{1-F^2} \right]^{\alpha \nu} $$
$$= \left( \partial_{\mu}  F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu} + 2  \left( \frac{F}{1 -F^2} \right)^{\mu \rho}   \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} \tag{2} $$
The second term in Eq.(2) cancels the second term in the RHS in the problem sheet equation. 
We then need to show
$$ \left( \partial_{\mu}  F^{\mu}_{\alpha} \right) \left( \frac{1}{1 -F^2} \right)^{\alpha \nu}  + \left( \frac{F}{1 -F^2} \right)^{\mu \rho}   \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{F}{1-F^2}\right)^{\sigma \nu} -  \left( \frac{1}{1 -F^2} \right)^{\mu \rho}   \left( \partial_{\mu} F \right)_{\rho\sigma} \left(\frac{1}{1-F^2}\right)^{\sigma \nu} =0 \tag{3} $$
then I didn't find a way to show Eq. (3) hold. I tried to combine the second and third term, and rearrange them, but didn't got a simple expression.
 A: Eq. ($\ast$) in David Tong's notes follows from the more general eq. ($\ast\ast$)
$$\tag{**}\begin{align}
\partial_{\mu}\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\nu}
&=\left(\frac{F}{1-F^2}\right)^{\lambda}{}_{\rho}
~\partial_{\mu}F^{\rho}{}_{\sigma}
\left(\frac{F}{1-F^2}\right)^{\sigma}{}_{\nu}
\\ &\qquad+\left(\frac{1}{1-F^2}\right)^{\lambda}{}_{\rho}
~\partial_{\mu} F^{\rho}{}_{\sigma}
\left(\frac{1}{1-F^2}\right)^{\sigma}{}_{\nu}  \end{align}$$
by putting the $\lambda$-index equal to the $\mu$-index, sum over $\mu$, and raise the $\nu$-index. Eq. ($\ast\ast$) is equivalent to eq. ($\ast\ast\ast$)
$$\tag{***}\begin{align}
\partial_{\mu}\left(\frac{F}{1-F^2}\right)
&=\left(\frac{F}{1-F^2}\right)
\partial_{\mu}F
\left(\frac{F}{1-F^2}\right)
\\ &\qquad+\left(\frac{1}{1-F^2}\right)
\partial_{\mu} F
\left(\frac{1}{1-F^2}\right), \end{align}$$
if we implicitly imply that every matrix $F$ carries one upper and one lower index a la $F^{\lambda}{}_{\nu}$. (Note that that there is no ambiguity in writing the matrix expression $\frac{F}{1-F^2}$ as a fraction because the numerator and the denominator commute.)
Finally, eq. ($\ast\ast\ast$)  follows by straightforward manipulations, which involve e.g. using the rule 
$$\tag{****} \partial_{\mu}\frac{1}{M}~=~-\frac{1}{M}\partial_{\mu}M\frac{1}{M} $$
of how to differentiate an inverse matrix $M$. (In particular, the Bianchi identity that David Tong mentions is not used in the proof of eq. ($\ast\ast\ast$).)
A: $\partial_i \frac{F}{1-F^2}=-(\frac{F}{1-F^2})\partial_i (\frac{1-F^2}{F})(\frac{F}{1-F^2})$
$=-(\frac{F}{1-F^2})\partial_i(\frac{1}{F}-F)(\frac{F}{1-F^2})$
$=-(\frac{F}{1-F^2})(\partial_i(\frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$
$=-(\frac{F}{1-F^2})(- \frac{1}{F}\partial_i F \frac{1}{F}) -\partial_iF)(\frac{F}{1-F^2})$
$=(\frac{1}{1-F^2})\partial_i F(\frac{1}{1-F^2}) + (\frac{F}{1-F^2})\partial_i F(\frac{F}{1-F^2})$
