       Re: Bypassing built-in functions in differentiation

• To: mathgroup at smc.vnet.net
• Subject: [mg62621] Re: Bypassing built-in functions in differentiation
• From: Peter Pein <petsie at dordos.net>
• Date: Tue, 29 Nov 2005 06:43:26 -0500 (EST)
• References: <dmh8hi\$8n7\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Ofek Shilon schrieb:
> Dear MathGroup.
>
> consider the following statement:
>
> Dt[Transpose[a]]
>
> which evaluates to
>
> Dt[a] Transpose`[a]
>
> that is, mathematica treats Transpose as a function and uses the chain
> rule. i can try and bypass this behaviour manually:
>
> Unprotect[Dt];
> Dt[Transpose[x_]] := Transpose[Dt[x]]
>
> but now consider expressions like -
>
> Dt[Transpose[a].b]
>
> which still produces:
>
> Transpose[a].Dt[b] + Dt[a] Transpose`[a] b
>
> which is a bit surprising. i can of course bypass this behaviour
> manually as well:
>
> Dt[Transpose[x_].y_] := Transpose[Dt[x]].y + Transpose[x].Dt[y]
>
> which gives the desired result, but then check the following -
> Dt[(Transpose[a].b)^2]
>
> etc. etc.
>
> i tried also to define -
> Dt[Transpose[x_]] =1
>
> which produces readable results, but discards the (correct, and needed)
> 'Transpose' head over a factor in the differentiation.
>
> There has to be a general solution. is there a 'hook' where i can
> interfere with the derivative computation? (i thought user definitions
> would suffice, but apparently not)
>
>
> thanks,
>
> Ofek
>
Hi Ofek,

dig a little deeper and change Transpose' :

In:= Unprotect[Transpose];
Derivative[Transpose] ^= 1 &
Out= 1&
In:= Dt[Transpose[a]]
Out= Dt[a]
In:= Dt[Transpose[a] . b]
Out= Dt[a].b+Transpose[a].Dt[b]
In:= Dt[(Transpose[a] . b)^2]
Out= 2 Transpose[a].b (Dt[a].b+Transpose[a].Dt[b])

etc.

Peter

```

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