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whatisthequestion
Group Title
Find the second derivaive of y = x / (x^3 1)
 one year ago
 one year ago
whatisthequestion Group Title
Find the second derivaive of y = x / (x^3 1)
 one year ago
 one year ago

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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
have you tried quotient rule or product rule + chain rule?
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
yes I got the first derivative correctly (verified by Wolfram Alpha) but am messing up when taking the derivative the second time
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
the first derivative is \[\frac{ 2x ^{3}+1 }{ (x ^{3}1)^{2} }\]
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.0
dw:1352709544762:dw
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
I know the quotient rule and chain rule but something is wrong with my execution of the rules
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.0
@whatisthequestion let me tell you a good one method. which is much better than this. Do you want to know??????? or want to try with method actually there is just calculation mistakes only:)
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
I am open to other methods if you have another method let me know
 one year ago

Aperogalics Group TitleBest ResponseYou've already chosen the best response.0
first make it into an equation. can u make.......try :)
 one year ago

jvromero Group TitleBest ResponseYou've already chosen the best response.2
\[y = \frac{ x }{ x^31 } = \frac{1}{x^2\frac{1}{x}}=\left[ x^2x^{1} \right]^{1}\] \begin{align} y' &= (1)\left[ x^2x^{1} \right]^{2} (2x  (1)x^{2}) \\ &= [ x^2x^{1} ]^{2} (2x + x^{2}) \\ &= \frac{2x+x^{2}}{(x^2  x^{1})^2}\\ &= \frac{2x^3+1}{x^2(x^2  x^{1})^2}\\ &= \frac{2x^3+1}{(x(x^2  x^{1}))^2}\\ &= \frac{2x^3+1}{(x^3  1)^2}\\ \end{align} \begin{align} y'' &= \frac{d}{dx}[\frac{2x^3+1}{(x^3  1)^2}]\\ &=\frac{d}{dx}[(x^3  1)^{2}(2x^3+1)]\\ &=\frac{d}{dx}[(x^3  1)^{2}](2x^3+1) + (x^3  1)^{2} \frac{d}{dx}[(2x^3+1)]\\ &= 2(3x^2)(x^3  1)^{3}(2x^3+1) + (x^3  1)^{2} (6x^2)\\ &= (6x^2)(2x^3+1)(x^3  1)^{3} + (6x^2)(x^3  1)^{2}\\ &= (12x^56x^2)(x^3  1)^{3} + (6x^2)(x^3  1)^{2}\\ &= \frac{12x^56x^2}{(x^3  1)^3} + \frac{6x^2}{(x^3  1)^2}\\ &= \frac{12x^56x^2}{(x^3  1)^3} + \frac{(6x^2)(x^3  1)}{(x^3  1)^3}\\ &= \frac{12x^56x^2 6x^5 +6x^2}{(x^3  1)^3}\\ &= \frac{6x^5}{(x^3  1)^3}\\ \end{align}
 one year ago

jvromero Group TitleBest ResponseYou've already chosen the best response.2
verification: https://www.wolframalpha.com/input/?i=differentiate+%28%E2%88%922x^3%2B1%29\%28x^3%E2%88%921%29^2+with+respect+to+x
 one year ago
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