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anonymous
 3 years ago
find the derivative of x(x1)^2
anonymous
 3 years ago
find the derivative of x(x1)^2

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my advice for you is..don't use product rule. expand (x1)^2 first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright so \[x^22x+1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why not to use product rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right. don't forget the x x(x^2  2x + 1) now distribute x into the trinomial

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SheldonEinstein because expanding it will be a lot simpler

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you expand it, you just get a polynomial (then you can derive the terms individually)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh! Right, I thought you were referring that "Product rule will not work here" . It's my bad, sorry :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then \[x^32x^2+x\] so f'(x) = \[3x^24x+1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you show product rule with that quickly please?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Though, I would like to show how product rule will work in this way : \[\large{\frac{d}{dx}{u.v} = u \frac{dv}{dx} + v \frac{du}{dx} \implies \textbf{Product rule}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0put u = x and v = (x1)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now? @sjerman1 can you do it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you required to solve it by product rule or are you just interested @sjerman1 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have no idea what that theorem is.....but let's see what @SheldonEinstein is typing

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for late answer :( I was checking it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Amazing job! Thank you very much!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're welcome @sjerman1 ... Thanks for the patience friends :) Also , good job @lgbasallote

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right ..... 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) ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i wonder, is there anyway to view these questions after they are closed?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Save the link of the question in your computer databse.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*database. This will help you to reopen it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or...you can just click your username on the upper right corner and then press questions asked in the profile page...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here the message says all.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There 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.
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