anonymous
  • anonymous
Find y" by implicit differentiation 9x^2 + y^2 =9
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
bye girl dont go posting questions on our chat log
anonymous
  • anonymous
Why can't I post it here?
amistre64
  • amistre64
what is your workings for y'?

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zepdrix
  • zepdrix
\(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\) Starting with a circle? Oo these are fun. Oh wait that's not a circle :d
anonymous
  • anonymous
i was joking
anonymous
  • anonymous
I got my y' as (-9x)/y And y" as (-9y-9xy')/y^2 I was trying to substitute y' back into y" But then I got (81xy+81x^2)/y^3 However the answer is ((9)(9x^2+y^2))/y^3
amistre64
  • amistre64
9x^2 + y^2 =9 9(2)x + 2y y' =0 9x + y y' =0 ; your y' is fine now derive again; notice that yy' is a product tho
anonymous
  • anonymous
And I don't understand why is there a y^2 on top
amistre64
  • amistre64
9x + y y' =0 9 + y'y' + yy'' = 0 yy'' = -(9 + y'^2) y'' = -(9 + y'^2)/y
DanJS
  • DanJS
y is a function of x, to take the derivative of y w.r.t. x, you have to use the chain rule
amistre64
  • amistre64
y'' = -(9 + y'^2)/y y'' = -(9 + (-9x/y )^2)/y y'' = -(9 + 81x^2/y^2)/y
amistre64
  • amistre64
multiply top and bottom by y^2 to clear the fraction in the numerator
DanJS
  • DanJS
\[\frac{ d }{ dx }y(x) = \frac{ d }{ dy }y(x)*\frac{ dy }{ dx }\]
zepdrix
  • zepdrix
`I got my y' as (-9x)/y` `And y" as (-9y-9xy')/y^2` These steps look correct ^ Maybe you just ran into a little trouble plugging in your y'. Hmm.
zepdrix
  • zepdrix
Oh there should be another negative in the numerator, I missed that. -(-9x)y'
zepdrix
  • zepdrix
It looks like you chose the quotient rule route, so remember that it's subtraction in the numerator, and we're carrying around a (-9x).
anonymous
  • anonymous
As my y' is (-9x/y) I used quotient rule to derive my y" That's y I came up with : Y"=(-9y-9xy')/y^2 y"= (-9y-9x)(-9x/y) / y^2 Y"= (81xy+81x^2)/y^3 However the answer shows: 9(9x^2+y^2)/y^3
amistre64
  • amistre64
using product (it tends to be simpler to follow) y' = -9xy^(-1) y'' = -9y^(-1) + 9xy'y^(-2) -9y^(-1) + 9x(-9xy^(-1))y^(-2) -9 y^(-1) -81x^2 y^(-3)
amistre64
  • amistre64
you sure the answer is without a negative?
anonymous
  • anonymous
There is a negative sign I'm sorry I missed it
zepdrix
  • zepdrix
\[\large\rm y''=\frac{-9y-(-9x)\color{orangered}{y'}}{y^2}=y''=\frac{-9y-(-9x)\color{orangered}{\frac{(-9x)}{y}}}{y^2}\]Depending on how you proceeded from here.... I would probably choose to factor a 1/y from each term in the numerator,\[\large\rm y''=\frac{1}{y}\cdot\frac{-9y^2+81x^2}{y^2}\]That last step is one giving you trouble? :o
zepdrix
  • zepdrix
I missed a negative again :) lol\[\large\rm y''=\frac{1}{y}\cdot\frac{-9y^2-81x^2}{y^2}\]
anonymous
  • anonymous
Ok I'm trying to figure it out right now. Thanks for all of your help:)!!!!!!!!!!!! Thanks!

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