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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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