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anonymous
 5 years ago
could someone work out y'' for y= ln x/x^2
anonymous
 5 years ago
could someone work out y'' for y= ln x/x^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{\ln x }{x^2} \rightarrow y'=\frac{x^2 (\ln x)'\ln x (x^2)'}{(x^2)^2}=\frac{x^2/x2x \ln x}{x^4}\]\[=\frac{x2x \ln x}{x^4}=\frac{x2\ln x}{x^3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i moved the x^2 to the top so i could use the product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y''= \left( \frac{x2 \ln x}{x^3} \right)'=\frac{x^3(x2 \ln x)'(x2\ln x)(x^3)'}{(x^3)^2}\]\[=\frac{x^3(12/x)(x2\ln x)(3x^2)}{(x^3)^2}\]\[=\frac{x^32x^23x^3+6x^2 \ln x}{x^6}=\frac{x23x+6\ln x}{x^4}\]\[=\frac{6\ln x2(x+1)}{x^4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ballards, your way is equivalent, and actually, my preferred method! I thought you might have needed it using the quotient rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0our teacher said use whatever way we can get it. could you work it out that way so i can check my work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm rushing through stuff. Hope it's what you've got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second line is the first derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your result seems fine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just need to simplify now your second derivative now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0so could you show me how to do that. i am stuck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0okay thank you so much!
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