Question

could someone work out y'' for y= ln x/x^2

Answer 1

here is the working

Answer 2

\[y=\frac{\ln x }{x^2} \rightarrow y'=\frac{x^2 (\ln x)'-\ln x (x^2)'}{(x^2)^2}=\frac{x^2/x-2x \ln x}{x^4}\]\[=\frac{x-2x \ln x}{x^4}=\frac{x-2\ln x}{x^3}\]

Answer 3

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

Answer 4

\[y''= \left( \frac{x-2 \ln x}{x^3} \right)'=\frac{x^3(x-2 \ln x)'-(x-2\ln x)(x^3)'}{(x^3)^2}\]\[=\frac{x^3(1-2/x)-(x-2\ln x)(3x^2)}{(x^3)^2}\]\[=\frac{x^3-2x^2-3x^3+6x^2 \ln x}{x^6}=\frac{x-2-3x+6\ln x}{x^4}\]\[=\frac{6\ln x-2(x+1)}{x^4}\]

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

here is what i got

Answer 9

the second line is the first derivative

Answer 10

Your result seems fine.

Answer 11

You just need to simplify now your second derivative now.

Answer 12

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.

Answer 13

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

Answer 14

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?

Answer 15

SIMPLIFYING

Answer 16

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.

Answer 17

okay thank you so much!