Question

Find y'' by implicit differentiation. 4x^2 + y^2 = 8.
I already got that y' = -4x/y but idk what to do next.

Answer 1

Please explain as you show steps. I get confused if all the steps arent included.

Answer 2

how did you get your y'?

Answer 3

4x^2 + y^2 = 8

what is the second derivative of 4x^2 with respect to x?

Answer 4

\(4x^2+y^2=8\\\frac{d}{dx}[4x^2+y^2]=\frac{d}{dx}[8]\\8x+2yy'=0\\4x+yy'=0\)

Now, we only need to implicitly differentiate *again*. We need to use the product rule to differentiate \(yy'\) to yield \(y'y'+yy''\), or \(y'^2+yy''\).

\(\frac{d}{dx}[4x+yy']=\frac{d}{dx}[0]\\4+y'^2+yy''=0\\yy''=-(4+y'^2)\\y''=-\frac{4+y'^2}y\)

Answer 5

I got 8x+2yy' = 0
2yy' = -8x
y' = -8x/ 2y -> -4x/y

Answer 6

Did i do that part correctly ??

Answer 7

8x + 2yy' = 0 is good

take the derivative of it again; notice the y parts is a product rule now

Answer 8

Why are you trying to isolate \(y'\)? Leave it \(8x+2yy'=0\) and you can just implicitly differentiate once more and then solve for \(y''\)!

Answer 9

divide off the 2 if need be or wait doesnt matter :)

Answer 10

y' is needed to "simplify" it after finding y''

Answer 11

If i differentiate it again it will be 8 + d/dx(2yy') ?

Answer 12

so far so good

Answer 13

So what should i do next amistre64?

Answer 14

spose we say 2y=a and y' = b

what is the derivative of: ab ? its just a product rule

Answer 15

@help_needed do you know how to take the derivative of \(2yy'\)?

Answer 16

remember ow to do a product?

Answer 17

\(\frac{d}{dx}[2y\frac{dy}{dx}]=2\frac{d}{dx}[y\frac{dy}{dx}]=2[\frac{dy}{dx}\frac{dy}{dx}+y\frac{d^2y}{dx^2}]=2[y'^2+yy'']\)

Answer 18

product rule is the same in all cases

say:

d/dx (ab) = a'b + ab'

d/dx(2yy') = 2 d/dx (yy') = 2(y'y'+yy'')

Answer 19

why does the y have a prime and a 2 as a superscript ?? Is that same as y''

Answer 20

y' * y' = (y')^2

Answer 21

8x + 2yy' = 0 ; lets divide off the 2, its just in the way

4x + yy' = 0 ; and diffy it

4 + (y' y' + y y')' = 0

Answer 22

got a bad ) in there :)

Answer 23

4 + (y' y' + y y'') = 0

4 + (y')^2 + y y'' = 0 ; and solve for y''

Answer 24

Is it y'' = -4- (y'^2) /y

Answer 25

let me chk

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

y y'' = -(4+ y'^2)

y'' = -(4+ y'^2)/y

and youve already determined what y' is to sub it in

Answer 26

i wish i could stay, but my real life job is needing my attention :) good luck

Answer 27

how come u didnt change the sig of y'^2 when bringing to other side. And its ok! Thanks for helping amistre64

Answer 28

You gave me the steps to solve which is what i really neeeded. Thanls afain @amistre64

Answer 29

Hopefully someone else can help me work thru the rst of the problem