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- anonymous

Find y" by implicit differentiation
9x^2 + y^2 =9

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- anonymous

Find y" by implicit differentiation
9x^2 + y^2 =9

- chestercat

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- anonymous

bye girl dont go posting questions on our chat log

- anonymous

Why can't I post it here?

- amistre64

what is your workings for y'?

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- 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

i was joking

- 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

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

And I don't understand why is there a y^2 on top

- amistre64

9x + y y' =0
9 + y'y' + yy'' = 0
yy'' = -(9 + y'^2)
y'' = -(9 + y'^2)/y

- 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

y'' = -(9 + y'^2)/y
y'' = -(9 + (-9x/y )^2)/y
y'' = -(9 + 81x^2/y^2)/y

- amistre64

multiply top and bottom by y^2 to clear the fraction in the numerator

- DanJS

\[\frac{ d }{ dx }y(x) = \frac{ d }{ dy }y(x)*\frac{ dy }{ dx }\]

- 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

Oh there should be another negative in the numerator, I missed that.
-(-9x)y'

- 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

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

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

you sure the answer is without a negative?

- anonymous

There is a negative sign I'm sorry I missed it

- 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

I missed a negative again :) lol\[\large\rm y''=\frac{1}{y}\cdot\frac{-9y^2-81x^2}{y^2}\]

- anonymous

Ok I'm trying to figure it out right now. Thanks for all of your help:)!!!!!!!!!!!! Thanks!

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