Problem - find second derivative

  • Thread starter jkristia
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  • #1
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The problem and my attempt at a solution is shown in the attached image.

The problem is that I end up with one extra x in the denominator.
So the question is: Is my expression for y'' correct and I just made a mistake somewhere (I have checked it several times), or am I missing something in the expression for y' and/or y'' ?

attachment.php?attachmentid=53971&stc=1&d=1355635814.png
 

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Answers and Replies

  • #2
MarneMath
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If the answer in the book is something like 6/x^3 + 18/x^4 + 12/x^5, you got the same answer.
 
  • #3
SammyS
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The problem and my attempt at a solution is shown in the attached image.

The problem is that I end up with one extra x in the denominator.
So the question is: Is my expression for y'' correct and I just made a mistake somewhere (I have checked it several times), or am I missing something in the expression for y' and/or y'' ?

[ IMG]https://www.physicsforums.com/attachment.php?attachmentid=53971&stc=1&d=1355635814[/PLAIN]
How do you go from [itex]\displaystyle\ \frac{6}{x^3}\left( 1+\frac{3}{x}+\frac{2}{x^2}\right)\ [/itex] to [itex]\displaystyle\ \frac{6}{x^4}\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ ? [/itex]

[itex]\displaystyle\ 1+\frac{3}{x}+\frac{2}{x^2}=\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ . [/itex]
 
  • #4
symbolipoint
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Does this make sense?
[tex] \[
\begin{array}{l}
y' = 3(1 + {\textstyle{1 \over x}})^2 x^{ - 2} ( - 1) \\
= \frac{{ - 3(1 + {\textstyle{1 \over x}})^2 }}{{x^2 }} \\
\end{array}
\]
[/tex]

Next you would use quotient rule to find the next derivative (second derivative).
 
  • #5
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How do you go from [itex]\displaystyle\ \frac{6}{x^3}\left( 1+\frac{3}{x}+\frac{2}{x^2}\right)\ [/itex] to [itex]\displaystyle\ \frac{6}{x^4}\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ ? [/itex]

Hmm, I probalby did a silly mistake, but I multiplied by x^2

[itex](\frac{6}{x^3})(\frac{x^2 +3x + 2}{x^2}) = (\frac{6}{x^3})(\frac{(x+1)(x+1)}{x^2})[/itex] to .... ahhh found my silly mistake - .. of course I need (x)(x), and then x can be cancelled - haha..

Thank you very much for you help.
 
  • #6
Curious3141
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Personally, I'd just expand the binomial then differentiate twice. Keep the exponents as negative powers, it makes everything quicker.
 
  • #7
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>>Personally, I'd just expand the binomial then differentiate twice
That certainly makes it much easier. I guess I didnt think of that since I was doing problems related to the chain rule. Thank you for your suggestion.
 

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