Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
my advice for you is..don't use product rule. expand (x1)^2 first
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
alright so \[x^22x+1\]
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Why not to use product rule?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
right. don't forget the x x(x^2  2x + 1) now distribute x into the trinomial
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
@SheldonEinstein because expanding it will be a lot simpler
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
if you expand it, you just get a polynomial (then you can derive the terms individually)
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Oh! Right, I thought you were referring that "Product rule will not work here" . It's my bad, sorry :)
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
then \[x^32x^2+x\] so f'(x) = \[3x^24x+1\]
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
can you show product rule with that quickly please?
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Though, I would like to show how product rule will work in this way : \[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
put u = x and v = (x1)^2
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Now? @sjerman1 can you do it?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
are you required to solve it by product rule or are you just interested @sjerman1 ?
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
i am just interested, that was a much easier way to solve it but I am actually in the process of finding points guaranteed to exist by rolle's theorem
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Wait I am typing :(
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
i have no idea what that theorem is.....but let's see what @SheldonEinstein is typing
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
\[\large{x \frac{d (x1)^2}{dx} + (x1)^2 \frac{dx}{dx} }\] \[\large{x \frac{d(x^2+12x)}{dx} + (x1)^2}\] \[\large{x [ \frac{d(x^2)}{dx} + \frac{d(1)}{d(x)}  \frac{d(2x)}{dx} ] +(x1)^2}\] \[\large{x [ 2x + 0  2 ] + x^2 +1  2x}\] \[\large{ 2x^2  2x + x^2 + 1  2x}\] \[\large{3x^24x+1}\]
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Sorry for late answer :( I was checking it :)
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
Amazing job! Thank you very much!
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
You're welcome @sjerman1 ... Thanks for the patience friends :) Also , good job @lgbasallote
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
you can also do it without expanding (x1)^2 in the derivative x(x1)^2 x[2(x1)] + (x1)^2 x[2x 2] + x^2  2x + 1 2x^2  2x + x^2  2x + 1 3x^2  4x + 1 just showing an alternative
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Right ..... but usually I prefer to just use d(x^n) / dx and let the students know that x can be anything. like x can be (x1) and similarly it can be (x+1) ...
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
i wonder, is there anyway to view these questions after they are closed?
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Yes @sjerman1
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Save the link of the question in your computer databse.
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
*database. This will help you to reopen it.
 2 years ago

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
Thank you both!
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
or...you can just click your username on the upper right corner and then press questions asked in the profile page...
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
Here the message says all.
 2 years ago

SheldonEinstein Group TitleBest ResponseYou've already chosen the best response.1
There in the message has "reopen" word but that means like you can not get in "OPEN QUESTIONS SECTION" again but still the other users can help you if they regularly see closed quest. section.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.