## anonymous one year ago Find y" by implicit differentiation 9x^2 + y^2 =9

1. anonymous

bye girl dont go posting questions on our chat log

2. anonymous

Why can't I post it here?

3. amistre64

what is your workings for y'?

4. 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

5. anonymous

i was joking

6. 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

7. 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

8. anonymous

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

9. amistre64

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

10. DanJS

y is a function of x, to take the derivative of y w.r.t. x, you have to use the chain rule

11. amistre64

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

12. amistre64

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

13. DanJS

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

14. 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.

15. zepdrix

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

16. 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).

17. 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

18. 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)

19. amistre64

you sure the answer is without a negative?

20. anonymous

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

21. 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

22. zepdrix

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

23. anonymous

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