Question

find d^2y/dx^2 of

x^2 + y^2 = 7

Answer 1

can you first find dy/dx ?

Answer 2

yeah x/y

Answer 3

dy/dx = - x/y

Answer 4

yeah i fixed what i did a second ago. i just dont know how to get the second part

Answer 5

as a first step, take the derivative of both sides of
\[ y' = -\frac{x}{y} \]

Answer 6

so the derivative of -x and -y?

Answer 7

use the quotient rule
or rewrite as
-1*x*y^(-1) and use the product rule

Answer 8

i still got -x/y

Answer 9

yes, now that we have
\[ \frac{dy}{dx} = - \frac{x}{y} \]
now take the derivative a second time:
\[ \frac{d^2y}{dx^2} = -\frac{d}{dx}\left( \frac{x}{y}\right) \]

Answer 10

quotient rule:
\[ d \frac{u}{v} = \frac{ v\ du - u\ dv}{v^2} \]

Answer 11

so the answer is -x^2+y2/y^2?

Answer 12

what did you get for d (-x/y) ?

Answer 13

i dont know... you said what the formula was and that was the answer choice that was closest

Answer 14

oh

we have to do this problem in steps. even after you find d (-x/y) there is one more step after that to get a "simplified" version of d^2 y/dx^2

you need to find d(-x/y)

Answer 15

ok to be honest i dont understand implicit differentistion at all

Answer 16

take 5 minutes and look over this post from yesterday
http://openstudy.com/users/thatguyintheback#/updates/52b4e4aae4b02932bbbfdb0b

then we will pick up here.

Answer 17

I assume if you had the problem
\[ \frac{d}{dx} \left(\frac{x}{\sin x}\right) \]
you could use the quotient rule ?
it is the same for
\[ \frac{d}{dx} \left(\frac{x}{y}\right) \]
we will use the fact that \( \frac{d}{dx} y= \frac{dy}{dx} \)

Answer 18

ok i have to eat breakfast but ill be on later and try to understand. ill work on it when im gone. thanks for your help!

Answer 19

ok, meanwhile
\[- \frac{d}{dx}\left(\frac{x}{y}\right) = -\left(\frac{ y \frac{d}{dx}x - x \frac{d}{dx}y}{y^2} \right) \\
-\left(\frac{ y - x \frac{dy}{dx}}{y^2} \right) \\
\left(\frac{x \frac{dy}{dx}-y}{y^2}\right)
\]

Answer 20

\[\huge y' = \frac{ -x }{ y }\]
\[\huge y''= \frac{ -y + xy' }{ y^2 }\]
Plug in y'.
\[\huge y'' = \frac{ -y + \frac{ -x^2 }{ y } }{ y^2 }\]
\[\huge y'' = \frac{ -y }{ y^2 } + (\frac{ -x^2 }{ y } * y^2)\]
\[\huge y'' = \frac{ -1 }{ y } -x^2y\]