## amy0799 one year ago Find d2^y/dx^2 in terms of x and y.

1. amy0799

$x ^{2}y ^{2}-8x=2$

2xy^2+x^2(2y)dy/dx-8=0

3. amy0799

I already have the first derivative figured out, I don't know how to find the second though

just change the sides take evrything to right side except x^2(2y)dy/dx

5. amy0799

do I just plug in x^2y^2−8x into the derivative of d2^y/dx^2?

nope u do like this |dw:1442671687613:dw|

do you understand

8. amy0799

umm not really, is that for dy/dx?

9. amy0799

how? I cant really read what you drew

okay both the drawings

11. amy0799

can you type out ur equations please?

wait

dy/dx=8/x^2y-2xy^2/x^22y dy/dx=4/x^2y-y/x

now do double derivative

15. amy0799

I got the first derivative to be $\frac{ xy ^{2}-4 }{ x^{2}y }$

nope i told the first one

this is not normal derivative this is implicit differentiation

18. amy0799

I know but when I did the implicit differentiation for x^2y^2−8x=2 I got that equation and it's different than yours

yeah i'm sorry dude you are correct

i'm so sorry actually i've simplified it a bit

21. amy0799

how did you simplify it?

22. amy0799

Hello?

forget that what you did is absolutely correct

but u made a mistake with the signs it will be 4-xy^2/x^2y

25. amy0799

oh ok. so I just find the derivative of that?

yes

27. amy0799

ok. ill show u what I get in a minute

no problem

29. amy0799

I got $\frac{ x ^{2}y(-2xyy'-y ^{2})-(4-xy ^{2})(x ^{2}y'+2xy) }{ y ^{2}x ^{4} }$ correct so far?

30. amy0799

r u there?

yes

32. amy0799

is what I had right so far?

yeah

anything more

35. amy0799

well I still haven't found the 2nd derivative.

36. amy0799

@IrishBoy123 can u help?

dude this is the second derivative

38. amy0799

ik but there should be more to it

see if you are able to simplify it more but this is f9

40. amy0799

f9?

yeah i mean fine

this is correct

43. amy0799

don't I need to plug in 4-xy^2/x^2y into dy/dx?

yes

45. amy0799

I got $\frac{ 2x^2 y^5-8xy^3-16}{ y}$ but that's the wrong answer when I summited it

46. IrishBoy123

your first deriv looks wrong way round for starters

47. amy0799

what is it suppose to be?

48. IrishBoy123

should be $\color{red} - \frac{ xy ^{2}-4 }{ x^{2}y }$

nope the first derivative done by amy is f9

i've checked that in wolfram alpha

51. IrishBoy123

you can then do it implicitly starting $y' .x^2 .y = 4 - xy^2$ and applying the triple product, see here |dw:1442677649641:dw| but, whatever way you do it, it's horrible.

????????

53. amy0799

@IrishBoy123 how would I find the answer?

y^2(y-3xy'(x))/x^2

56. IrishBoy123

first deriv $x ^{2}y ^{2}-8x=2$ $2x y ^{2} + x^22yy'-8 = 0$ $y' = \frac{8-2xy^2}{2x^2y} = ....$ie wrong for second deriv you **might** go: $(x^2)(y)(y')=4 -x y ^{2}$ $(x^2)'(y)(y') +(x^2)(y)'(y')+(x^2)(y)(y')' = -y^2-2xyy'$ at some point you plug in for y'

57. amy0799

isn't that gonna be a big number when I plug in y'?

58. IrishBoy123

it's going to be horrible whatever way you do it. i have a feeling this could be a more organised way to do it, but if you are most of the way through, carry on as before however, do check that first derivative again or you will never get there