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

1. anonymous

here is the working

2. 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}$

3. anonymous

i moved the x^2 to the top so i could use the product rule

4. 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}$

5. anonymous

ballards, your way is equivalent, and actually, my preferred method! I thought you might have needed it using the quotient rule.

6. anonymous

our teacher said use whatever way we can get it. could you work it out that way so i can check my work?

7. anonymous

8. anonymous

I'm rushing through stuff. Hope it's what you've got.

9. anonymous

here is what i got

10. anonymous

11. anonymous

the second line is the first derivative

12. anonymous

13. anonymous

You just need to simplify now your second derivative now.

14. 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.

15. anonymous

so could you show me how to do that. i am stuck

16. 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?

17. anonymous

SIMPLIFYING

18. 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.

19. anonymous

okay thank you so much!

20. anonymous