anonymous
  • anonymous
could someone work out y'' for y= ln x/x^2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
here is the working
anonymous
  • anonymous
\[y=\frac{\ln x }{x^2} \rightarrow y'=\frac{x^2 (\ln x)'-\ln x (x^2)'}{(x^2)^2}=\frac{x^2/x-2x \ln x}{x^4}\]\[=\frac{x-2x \ln x}{x^4}=\frac{x-2\ln x}{x^3}\]
anonymous
  • anonymous
i moved the x^2 to the top so i could use the product rule

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anonymous
  • anonymous
\[y''= \left( \frac{x-2 \ln x}{x^3} \right)'=\frac{x^3(x-2 \ln x)'-(x-2\ln x)(x^3)'}{(x^3)^2}\]\[=\frac{x^3(1-2/x)-(x-2\ln x)(3x^2)}{(x^3)^2}\]\[=\frac{x^3-2x^2-3x^3+6x^2 \ln x}{x^6}=\frac{x-2-3x+6\ln x}{x^4}\]\[=\frac{6\ln x-2(x+1)}{x^4}\]
anonymous
  • anonymous
ballards, your way is equivalent, and actually, my preferred method! I thought you might have needed it using the quotient rule.
anonymous
  • anonymous
our teacher said use whatever way we can get it. could you work it out that way so i can check my work?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I'm rushing through stuff. Hope it's what you've got.
anonymous
  • anonymous
here is what i got
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
the second line is the first derivative
anonymous
  • anonymous
Your result seems fine.
anonymous
  • anonymous
You just need to simplify now your second derivative now.
anonymous
  • anonymous
The result in my attachment is what you should end up with. Ignore what I typed up on this thing earlier. Working on this thing is impossible.
anonymous
  • anonymous
so could you show me how to do that. i am stuck
anonymous
  • anonymous
It's kind of difficult to explain much further online :( Are you having a problem with the working in my attachment, or in simplifying the algebra in your own answer?
anonymous
  • anonymous
SIMPLIFYING
anonymous
  • anonymous
I would first multiply each of your products to get four little expressions, and then collect like terms. I started typing but it's going to take forever. I will do it on paper and scan.
anonymous
  • anonymous
okay thank you so much!
anonymous
  • anonymous

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