Question

Find y'' by implicit differentiation
5x^2+y^2=6

Answer 1

Welcome to OpenStudy! :)

Answer 2

Thanks :)

Answer 3

1. apply the derivative operator (d/dx) to each of the three terms in the original equation. Note that we regard y as a function of x, and therefore must use the chain as well as the power rule in finding the derivative with respect to x of y^2.
2. Solve the resulting equation for the first derivative, y '(x).
3. Apply the derivative operator to y '(x),
4. Simplify the result.
Good luck! I'm loggiing off, but hope this helps you get started!

Answer 4

no wait

Answer 5

Label your final answer as y ''(x).

Answer 6

i got the first derivative -5x/y

Answer 7

could you help me with the second derivative?

Answer 8

you found y', but its easier (for me) to play with y'' format if you retain the implicit sturcture

5x^2 + y^2 = 6

5x + y y' = 0 derive again

Answer 9

\[5x^2+y^2=6,5*2x+2yy'=0,5x+yy'=0\]

Answer 10

Be certain to label that first derivative! I agree that it may be faster to find the second derivative by leaving the result of the first differentiation in implicit form. Note that this requires that you use the PRODUCT rule, as well as the CHAIN rule.

Answer 11

i cant seem to get it right.. probably making some mistake in between

Answer 12

type out your steps :)

Answer 13

\[\Large \underbrace{5x}_{power~rule} + \underbrace{y y'}_{product/chain} = \underbrace{0}_{constant}\]

Answer 14

is y'=-5x/y ?

Answer 15

yes, y' = -5x/y is good to keep in mind

Answer 16

5x+y.(-5x/y)=0 ?

Answer 17

or quotient rule:\[\frac tb\to\frac{bt'-b't}{b^2}\]

Answer 18

not sure if we should subsititute y' in the middle of it, i tend to do it at the end.

5x + y y' = 0

5 + [y' y' + y y''] = 0

Answer 19

so 5 + [(-5x/y)(-5x/y)+(y)(y'')=0 ? but i don't know what y'' is

Answer 20

y'' is some unknown that we algebra, in the same way you algebraed y'

Answer 21

5 + y'y' + y y'' = 0

5 + (y')^2 + y y'' = 0

y y'' = -(5+(y')^2)

y'' = -(5+(y')^2)/y

Answer 22

how much you want to simplify it is up to you :)

Answer 23

thanks.. i'm trying to get the correct answer

Answer 24

\[y'=\frac{-5x}{y}~~\to ~~y''=\frac{-5y+5xy'}{y^2}\]
the correct answer depends on how far you want to simplify it

Answer 25

what should i substitute with y?

Answer 26

y is just y ... just as x is just x. they tend to be place holders for some (x,y) point from the original function

Answer 27

in other words, an implicit derivative maintains the possibility that any derivative is a function of x and y

Answer 28

just as y' = -5x/y is a function in x and y, for some known (x,y) point that satisfies the original setup

Answer 29

-30/y^3 ??

Answer 30

maybe .. lets see.

\[y'' = -\frac{5+(y')^2}{y}\]
\[y'' = -\frac{5+(-5\frac xy)^2}{y}\]
\[y'' = -\frac{5+25\frac {x^2}{y^2}}{y}\]
\[y'' = -\frac{5y^2+25x^2}{y~y^2}\]
\[y'' = -5\frac{y^2+5x^2}{y^3}\]

Answer 31

or

\[y'=\frac{-5x}{y}~~\to ~~y''=\frac{-5y+5xy'}{y^2}\]
\[y''=\frac{-5y+5x(-5\frac xy)}{y^2}\]
\[y''=\frac{-5y-25\frac{x^2}{y}}{y^2}\]
\[y''=\frac{-5y^2-25x^2}{y^2~y}\]
\[y''=-5\frac{y^2+5x^2}{y^3}\]

Answer 32

the correct answer is -30/y^3, is there any way u can get that answer from the value of y'' above?

Answer 33

??

Answer 34

where are you getting that 'correct' answer from?

Answer 35

and no, i have shown 2 ways to approach the correct answer and they both match. so this -30/y^3 seems suspect to me

Answer 36

i entered the answer according to your explanation for y'' but it is wrong, i'm doing it online and have used all 3 of my submissions so it gives the answer after no more tries are left

Answer 37

i think there is some other way

Answer 38

can you post the entire question, i think you are mixing information here or leaving something out.

Answer 39

even the wolf agrees with my assessment

http://www.wolframalpha.com/input/?i=%285+%28x+%28-5x%2Fy%29-y%29%29%2Fy%5E2

Answer 40

i mean i know i'm not gonna get the points back for that question but i still want to know the right one

Answer 41

i know i totally understand your explanation

Answer 42

take a screen shot, and attach it please

Answer 43

but i don't know what the question wants

Answer 44

i'm working on the second question now

Answer 45

screen shot that one then, lets take the middle man out of the setup :)

Answer 46

i dont know how to :/

Answer 47

but seriously the answer is -30/y^3 and questions is asking to find y'' by implicit differentiation

Answer 48

on your keyboard is a button that youve never used, PrtScrn, its aroung hte scroll lock button .... press it.

open a paint program and paste it in. save the file and attach it

Answer 49

y' = -5x/y is correct, so the y'' is gonna have to have some sort of x in it, so i have to wonder about the content of the question, or the validity of the grading program :)

Answer 50

okay i know this questions is confusing me a lot

Answer 51

your input was off, even tho the 'answer' is wrong by a long shot lol

\[\large y''=-5~\frac{y^2+5x^2}{y^3}\]
or
\[\large y''=\frac{-5y^2-25x^2}{y^3}\]
NOT
\[\large y''=-5y-\frac{25x^2}{y^2}\]

Answer 52

but yeah ... thats messed up regardless lol

Answer 53

i know i did make a mistake but that was before you showed me the right way lol

Answer 54

i know right

Answer 55

show it to your teacher .... if the 1pt is that important ;)

Answer 56

so what do i do ???? i have more questions like this

Answer 57

i have like 8 questions on this topic and that 1 point constitutes to about 13% of the whole'

Answer 58

lol

Answer 59

brb in 2 mins... dont go!!

Answer 60

y' = -5x/y is good

or in product form

y' = -5x y^(-1)

y'' = -5y^(-1) + 5xy' y^(-2)

substitute in y'

y'' = -5y^(-1) + 5x(-5x/y) y^(-2)

y'' = -5/y -25x^2/y^3

y'' = [-5y^2 - 25x^2 ]/y^3

Answer 61

ill stay here, gonna do some desk work

Answer 62

now that you know how to screenshot, feel free to attach another one

Answer 63

back.. and yeah thanks so much for that :)

Answer 64

ok so here's another one
its super similar to the previous one lol

Answer 65

then lets make a general rule :)

ax^2 + by^2 = c

2ax x' + 2b y y' = 0 ; divide off the 2s, and x' = dx/dx = 1

ax + by y' = 0, y' = -ax/by

continue with y'' ....

ax + by y' = 0

ax' + by' y' + by y'' = 0; x'=1 again

a + b(y')^2 + by y'' = 0

by y'' = -a - b(y')^2

y'' = [-a - b(y')^2 ] / by

plug in y'

y'' = [-a - b(-ax/by)^2 ] / by

y'' = [-a - b(a^2x^2/b^2y^2) ] / by

y'' = [-a - a^2x^2/by^2 ] / by

y'' = [-aby^2 - a^2x^2 ] / [b^2 y^3]

Answer 66

so, a=4, b=1, c=3

y'' = [-4y^2 - 16x^2 ] / [y^3]

or as last time, a=5, b=1, c= whocares its not in the end

y'' = [-5y^2 - 25x^2 ] / [y^3]

Answer 67

now, according to the correctness of the last one, the solution HAS TO be -20/y^3 lol

Answer 68

let me solve it myself.. give me minute

Answer 69

k :)

Answer 70

lol how did u come up with the solution of -20/y^3?

Answer 71

well, since -5-25 = -30 i just assume their ignorance and let -4-16 = -20

Answer 72

that isn't correct though lol

Answer 73

the correct answer is ....
\[\large y'' = \frac{-4y^2 - 16x^2 }{y^3}\]

Answer 74

what they want the correct answer to be is anyones guess :)

Answer 75

theres a watch it option, click it to see what happens

Answer 76

i got it wrong again... it was -12/y^3

Answer 77

the 'watch it' option just takes me to a video where a guy is solving for y' but not y''

Answer 78

x^2 y^2
---- + --- = 1
3/4 3/1



at the point 0,sqrt(3)

y' = -4(0)/sqrt(3) = 0 as expected

y'' = (-4(0)-16(3))/(3sqrt(3))

y'' = -16/sqrt(3), which is the rate at which the slope is changing