Dsy Network www | forum | my | didattica | howto | wiki | el goog | stats | blog | dona | rappresentanti
Homepage
 Register   Calendar   Members  Faq   Search  Logout 
.dsy:it. : Powered by vBulletin version 2.3.1 .dsy:it. > Didattica > Corsi A - F > Calcolo delle probabilità e statistica matematica > problema con derivata..
  Last Thread   Next Thread
Author
Thread    Expand all | Contract all    Post New Thread    Post A Reply
Collapse
homerfdl
..::..

User info:
Registered: Mar 2005
Posts: 117 (0.02 al dì)
Location:
Corso:
Anno:
Time Online: 18:12:06: [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged
Thumbs up problema con derivata..

nellesame 17/2/05
esercizio 2 parte 1 si chiede il calcolo della densita di probabilita

densitaD=d/dx (1-e^(-vx))
qualcuno mi rinfresca come si calcola questa derivata passo per passo fino ad arrivare alla soluzine ve^-vx......
grazie....

04-04-2007 16:02
Click Here to See the Profile for homerfdl Find more posts by homerfdl Add homerfdl to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
Collapse
imperator
.consigliere.

User info:
Registered: Apr 2004
Posts: 146 (0.02 al dì)
Location:
Corso:
Anno:
Time Online: 5 Days, 6:58:26 [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged

ciao, provo a spiegartelo:
1) la derivata di una costante è 0
2) la derivata di e^f(x) è = a
f '(x) * e^f(x)
3) sia h(x) = f(x) - g(x); allora h'(x) = f '(x) - g'(x)

dunque:
1)f '(1)=0;

2)posto f(x) = -vx, ho che
la derivata di e^-vx é = a:
f '(-vx) * e^-vx , cioè
-v * e^-vx

3)la derivata di 1-e^(-vx) è quindi:
f '(1) - (-v * e^-vx) = 0 - (-v * e^-vx) = - (-v * e^-vx) = ve^-vx

spero di essere stato abbastanza chiaro, in ogni caso se hai bisogno chiedi pure

Last edited by imperator on 04-04-2007 at 16:37

04-04-2007 16:31
Click Here to See the Profile for imperator Click here to Send imperator a Private Message Find more posts by imperator Add imperator to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
Collapse
homerfdl
..::..

User info:
Registered: Mar 2005
Posts: 117 (0.02 al dì)
Location:
Corso:
Anno:
Time Online: 18:12:06: [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged

tutto chiarissssssimo!!!!!
grazie mille!!!!!:D:D

04-04-2007 16:41
Click Here to See the Profile for homerfdl Find more posts by homerfdl Add homerfdl to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
Collapse
imperator
.consigliere.

User info:
Registered: Apr 2004
Posts: 146 (0.02 al dì)
Location:
Corso:
Anno:
Time Online: 5 Days, 6:58:26 [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged

de nada

04-04-2007 16:48
Click Here to See the Profile for imperator Click here to Send imperator a Private Message Find more posts by imperator Add imperator to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
Collapse
omnibusy
Dottore

User info:
Registered: Jan 2004
Posts: 220 (0.03 al dì)
Location: Roma
Corso: informatica F1Y
Anno: 1 anno
Time Online: 1 Day, 19:53:03: [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged

visto che si parla di derivate ho un quesito:
a pag 99 del mood la derivata prima di mx(t) = (pe^t+q)^n è (npe^t)(pe^t+q)^(n-1) fin qui perfetto perchè per t=0 diventa mx'(t) = np.
il libro però calcola la derivata seconda così mx"(t) = n(n-1)(pe^t)^2(pe^t+q)^(n-2) + (npe^t)(pe^t+q)^(n-1) ma a me viene n(n-1)(pe^t)(pe^t-1)(pe^t+q)^(n-2) + (npe^t)(pe^t+q)^(n-1)
in sostanza al posto di (pe^t)^2 mi viene (pe^t)(pe^t-1) .
mi sapete dire dove sbaglio?

__________________
Se conosci il tuo nemico e conosci te stesso non temi l'esito di cento battaglie. Sun Tzu.

10-04-2007 15:08
Click Here to See the Profile for omnibusy Click Here to See the Blog of omnibusy Click here to Send omnibusy a Private Message Visit omnibusy's homepage! Find more posts by omnibusy Add omnibusy to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
Collapse
omnibusy
Dottore

User info:
Registered: Jan 2004
Posts: 220 (0.03 al dì)
Location: Roma
Corso: informatica F1Y
Anno: 1 anno
Time Online: 1 Day, 19:53:03: [...]
Status: Offline

Post actions:

Edit | Report | IP: Logged

scusate ho realizzato solo adesso.

__________________
Se conosci il tuo nemico e conosci te stesso non temi l'esito di cento battaglie. Sun Tzu.

10-04-2007 15:18
Click Here to See the Profile for omnibusy Click Here to See the Blog of omnibusy Click here to Send omnibusy a Private Message Visit omnibusy's homepage! Find more posts by omnibusy Add omnibusy to your buddy list Printer Friendly version Email this Article to a friend Reply w/Quote
All times are GMT. The time now is 02:22.    Post New Thread    Post A Reply
  Last Thread   Next Thread
Show Printable Version | Email this Page | Subscribe to this Thread | Add to Bookmarks

Forum Jump:
Rate This Thread:

Forum Rules:
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is OFF
vB code is ON
Smilies are ON
[IMG] code is ON
 

Powered by: vBulletin v2.3.1 - Copyright ©2000 - 2002, Jelsoft Enterprises Limited
Mantained by dsy crew (email) | Collabora con noi | Segnalaci un bug | Archive | Regolamento | Licenze | Thanks | Syndacate
Pagina generata in 0.054 seconds (64.86% PHP - 35.14% MySQL) con 25 query.