Find d2y/dx2 by implicit differentiation, given
x^3 + y^3 = 6.
Please help!

Question

Find d2y/dx2 by implicit differentiation, given 
x^3 + y^3 = 6.
Please help!

amistre64 · Answer

how is implicit any different then unimplicit?

amistre64 · Answer

what the derivative of x^3?

anonymous · Answer

3x^2

amistre64 · Answer

close, but you threw out your derived bit .... bad bad bad

3x^2 x'

amistre64 · Answer

i never said to derive it with respect to x

amistre64 · Answer

what is the derivative of y^3?

anonymous · Answer

you don't need the x'?

amistre64 · Answer

there is nothing special about an x ....

amistre64 · Answer

what is the derivative of y^3  :)

anonymous · Answer

and the derivative of y^3 derived implicitly is 3y^2 dy/dx or 3y^2 y'

amistre64 · Answer

right; 3y^2 y'

amistre64 · Answer

soo ...

3x^3 x' + 3y^2 y' = 0

agreed?

amistre64 · Answer

err..  3x^2 lol

anonymous · Answer

haha yes

anonymous · Answer

I know how to get the first derivative, it ends up being -(x^2)/(y^2)

anonymous · Answer

but for some reason I cannot get the 2nd derivative...

hero · Answer

Amistre, should I just post wolfram's version on here?

amistre64 · Answer

now lets clean it up;  with respect to x; x'=1 sooo

3x^2 + 3y^2 y' = 0 ; and solve for y'

3y^2 y' = -3x^2

y' = -3x^2/3y^2 ; and simplify

y' = -x^2/y^2 ; and simplify

amistre64 · Answer

if you want, its a free post ;)

hero · Answer

I agree. It's free of charge

amistre64 · Answer

you know the quotient rule right?

anonymous · Answer

Yes I do... I mean I've tried it and I apparently don't arrive at the right answer, so I'm probably not doing something right. That's why I need help! haha

amistre64 · Answer

lets see where I mess it up at then ;)

hero · Answer

gn, what grade are you in?

anonymous · Answer

why does that matter? lol

anonymous · Answer

but I have a meeting to go to that starts at 7! If you guys could help me with this problem real quick I would greatly appreciate it!!!

amistre64 · Answer

$$y' = \frac{-x^2}{y^2}$$
$$y' = \left(\frac{-x}{y}\right)^2$$
$$y'' = 2\left(\frac{-x}{y}\right)*\frac{d}{dx}\frac{-x}{y}$$
$$y'' = 2\left(\frac{-x}{y}\right)*\frac{-x'y+xy'}{y^2}$$
so far so good?

hero · Answer

show off

amistre64 · Answer

lol  im a genius

anonymous · Answer

yesss

amistre64 · Answer

$$y'' = 2\left(\frac{-x}{y}\right)*\frac{-x'y+xy'}{y^2}$$
$$y'' = \frac{-2x}{y}\frac{(-y+xy')}{y^2}$$
$$y'' = \frac{-2x(-y+xy')}{y^3}$$
and recall what y' was .... -x^2/y^2 right?

$$y'' = \frac{-2x(-y+x\frac{-x^2}{y^2})}{y^3}$$

hero · Answer

I have no idea how he types all of that so fast.

hero · Answer

Some secret program

anonymous · Answer

this all looks good and makes sense so far

amistre64 · Answer

$$y'' = \frac{2x(y+x\frac{x^2}{y^2})}{y^3}$$
$$y'' = \frac{2x(y^3+x^3)}{y^5}$$
$$y'' = \frac{2xy^3+2x^4}{y^5}$$
and i think that as simple as it goes

amistre64 · Answer

whatd you come up with?

anonymous · Answer

I came up with that except I I don't get how you go from that first step to the second one... so how does (y+x (x^2/y^2)) become (y^3+x^3)? Probably a dumb question but I don't see how that works

amistre64 · Answer

x*x^2 = x^3  right?

amistre64 · Answer

then to add fraction you need like denominators like they teach you back in kindergarten; so y^2 it up

amistre64 · Answer

y*y^2 = y^3

amistre64 · Answer

the denom of it sinks under to merge into the y^3

anonymous · Answer

I just put that answer in and it was wrong :(

amistre64 · Answer

..... then it prolly aint as simplified as it wants

hero · Answer

amistre's getting rusty

amistre64 · Answer

rusty?  i mess up all the time, this is par for the course :)

amistre64 · Answer

wolfram aint computing my input right

anonymous · Answer

So for a similar problem: Find d2y/dx2 by implicit differentiation, given 
x^3 + y^3 = 8. the answer is -16x/(y^5)

amistre64 · Answer

yeah, looks like i got the y^5 part

amistre64 · Answer

the first still goes to -x^2/y^2

amistre64 · Answer

$$y'=\frac{-x^2}{y^2}$$
$$y''=\frac{y^2(-x^2)'-(y^2)'(-x^2)}{y^4}$$
$$y''=\frac{y^2(-2x)-(2y\ y')(-x^2)}{y^4}$$
$$y''=\frac{-2xy^2+(2x^2y\ y')}{y^4}$$

amistre64 · Answer

$$y''=\frac{-2xy^2+(2x^2y\ \frac{-x^2}{y^2})}{y^4}$$
$$y''=\frac{-2xy^2+(2x^2\ \frac{-x^2}{y})}{y^4}$$
$$y''=\frac{-2xy^2-(2x^2\ \frac{x^2}{y})}{y^4}$$
$$y''=\frac{-2xy^2-(2x^4\ \frac{1}{y})}{y^4}$$
$$y''=\frac{-2xy^2*\frac{y}{y}-(2x^4\ \frac{1}{y})}{y^4}$$
$$y''=\frac{-2xy^3\frac{1}{y}-2x^4\ \frac{1}{y}}{y^4}$$
$$y''=\frac{-2xy^3-2x^4}{y^5}$$
hmmmm

amistre64 · Answer

what does y^3=?

anonymous · Answer

Sorry... I am really late for my meeting now... I can't stay on here any longer. Thank you so much for your help

amistre64 · Answer

x^3 + y^3 = 8

y^3 = 8-x^3 ..right?

amistre64 · Answer

$$y''=\frac{-2x(8-x^3)-2x^4}{y^5}$$
$$y''=\frac{-16x+2x^4-2x^4}{y^5}$$
$$y''=\frac{-16x}{y^5}$$

amistre64 · Answer

that did it

amistre64 · Answer

it wasnt that it was wrong, it just wasnt as simplified as they wanted it to be