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MathGroup Archive 2004

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Re: Forcing a Derivative

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50776] Re: [mg50753] Forcing a Derivative
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 19 Sep 2004 21:39:53 -0400 (EDT)
  • References: <200409190756.DAA17973@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

I thought the answer was easy, but the more I looked into it, the more strange behavior I found. For instance, I expected part of the problem to be that Derivative[2][f * g] doesn't mention x, and f*g isn't a pure function. The same applies to these derivatives, but they both work:

f[x_]=x^2+7;
g[x_]=3x^3+23;
Derivative[2][f]
Derivative[2][g]

2&

18 #1&

This one works but works strangely:

prod=Function[x,f[x]g[x]];
Derivative[2][prod]

Function[x, 18*x*f[x] + 2*g[x] + 2*Derivative[1][f][x]* Derivative[1][g][x]]

Mathematica knows f and g or it couldn't have derived that answer; yet it leaves much of it unevaluated. It is correct as far as it goes, however:

Derivative[2][prod][x];
D[f[x]*g[x], {x, 2}] // Expand
% == %% // Simplify

46 + 126*x + 60*x^3
True

Bobby

On Sun, 19 Sep 2004 03:56:01 -0400 (EDT), Scott Guthery <sguthery at mobile-mind.com> wrote:

> How does one force Derivative[n] to actually take the derivative?
>
> For example if ...
>
> f[x_] = x^2 + 7
>
> g[x_]=3x^3 + 23
>
> then
>
> Derivative[2][f * g]
>
> just puts a couple of primes on the product rather than actually computing the dervative.
>
> Thanks for any insight.
>
> Cheers, Scott
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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