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Re: Bypassing built-in functions in differentiation


The 'HoldPattern' and 'if' condition you suggested do seem to solve the
issues i stated - however, i noticed a worse problem:
The original hack (Transpose'[x_]:=Transpose[Dt[x]] / Dt[x]  ) is
completely unextendable to differentiation by an explicit variable.
That is, how can we obtain with it something like

In[56815]= Dt[Transpose[a].b , x]
Out[56815]= Transpose[Dt[a,x]].b + Transpose[a].Dt[b,x]

?

any ideas?

thanks,

Ofek


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