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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0bye girl dont go posting questions on our chat log

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why can't I post it here?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is your workings for y'?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\) Starting with a circle? Oo these are fun. Oh wait that's not a circle :d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got my y' as (9x)/y And y" as (9y9xy')/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
 one year ago
Best ResponseYou've already chosen the best response.19x^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
 one year ago
Best ResponseYou've already chosen the best response.0And I don't understand why is there a y^2 on top

amistre64
 one year ago
Best ResponseYou've already chosen the best response.19x + y y' =0 9 + y'y' + yy'' = 0 yy'' = (9 + y'^2) y'' = (9 + y'^2)/y

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0y is a function of x, to take the derivative of y w.r.t. x, you have to use the chain rule

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y'' = (9 + y'^2)/y y'' = (9 + (9x/y )^2)/y y'' = (9 + 81x^2/y^2)/y

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1multiply top and bottom by y^2 to clear the fraction in the numerator

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx }y(x) = \frac{ d }{ dy }y(x)*\frac{ dy }{ dx }\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0`I got my y' as (9x)/y` `And y" as (9y9xy')/y^2` These steps look correct ^ Maybe you just ran into a little trouble plugging in your y'. Hmm.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Oh there should be another negative in the numerator, I missed that. (9x)y'

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0As my y' is (9x/y) I used quotient rule to derive my y" That's y I came up with : Y"=(9y9xy')/y^2 y"= (9y9x)(9x/y) / y^2 Y"= (81xy+81x^2)/y^3 However the answer shows: 9(9x^2+y^2)/y^3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1using 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
 one year ago
Best ResponseYou've already chosen the best response.1you sure the answer is without a negative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There is a negative sign I'm sorry I missed it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm y''=\frac{9y(9x)\color{orangered}{y'}}{y^2}=y''=\frac{9y(9x)\color{orangered}{\frac{(9x)}{y}}}{y^2}\]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
 one year ago
Best ResponseYou've already chosen the best response.0I missed a negative again :) lol\[\large\rm y''=\frac{1}{y}\cdot\frac{9y^281x^2}{y^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I'm trying to figure it out right now. Thanks for all of your help:)!!!!!!!!!!!! Thanks!
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