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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
 one year ago

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

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

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

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

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
can you show product rule with that quickly please?
 one year 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 : \[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]
 one year ago

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

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

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

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

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

sjerman1 Group TitleBest ResponseYou've already chosen the best response.1
Amazing job! Thank you very much!
 one year 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
 one year 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
 one year 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) ...
 one year 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?
 one year ago

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

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

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

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

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