       Re: Backwards?

• To: mathgroup at smc.vnet.net
• Subject: [mg3518] Re: [mg3480] Backwards?
• From: jpk at apex.mpe.FTA-Berlin.de (Jens-Peer Kuska)
• Date: Wed, 20 Mar 1996 02:49:46 -0500
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Artur,

>
> I would like to ask a question please.
> How can I let Mathematica do this:
>
> 1) c[x,y]*a[x]*D[b[x],x]+c[x,y]*b[x]*D[a[x],x]+d[x,y]
>    --->c[x,y]*D[a[x]*b[x],x]+d[x,y]  ?
>    It's backward-chainrule.

the following works nice
(* create the formula *)
In[]=
Expand[
c[x,y]*D[ a[x]*b[x],x]
] +d[x,y]

Out[]=
d[x, y] + b[x] c[x, y] a'[x] +

a[x] c[x, y] b'[x]

(* show the rule for multiplication more explicitly *)

In[]=
Collect[%,c[x,y]]

Out[]=
d[x, y] + c[x, y]

(b[x] a'[x] + a[x] b'[x])

(* revert the rule for products of derivatives *)

In[]=
% /. any_. f1_[x_]*Derivative[n_.+1][f2_][x_]+
any_. f2_[x_]*Derivative[n_.+1][f1_][x_] :>
any*Derivative[n+1][f1*f2][x]

Out[]=

d[x, y] + c[x, y] (a b)'[x]

>
>
> 2) If I let u^2+v^2:=1 (sic), how cann I do:
>
>    a*u^2+a*v^2==a*1==a   ?

eqn=u^2+v^2==1;

Expand[a*#] & /@ eqn

gives the desired result.

Hope that helps
Jens

PS: The chain rule for derivatives is
D[f[g[x]],x] :> Derivative[f][g[x]]*Derivative[g][x] or

d ( f(g(x)) )
------------- = f'(g(x)) g'(x)
d x

You want reverse the rule for the derivatives of products.

Jens

==== [MESSAGE SEPARATOR] ====

```

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