Verify:

x^2 + y^2 = 1 Take the derivative of both sides
2x + 2y*y' = 0
y' = -x/y Find the second derivative

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

Question

Verify:

x^2 + y^2 = 1     Take the derivative of both sides
2x + 2y*y' = 0
y' = -x/y                  Find the second derivative

y'' = (-y + xy')/y^2

anonymous · Answer

I think u r correct...

baldymcgee6 · Answer

Wolfram says y'' = -1/y

anonymous · Answer

y'' = (-y + xy')/y^2
Wolfram says y'' = -1/y
Substitute y' = -x/y
y'' = (-y + x(-x/y))/y^2
y'' = -(x^2+y^2)/y^3
substitute x^2 = 1-y^2
simplify, y'' = -1/y^3 is what I got . . .   :-/

baldymcgee6 · Answer

hmmm

anonymous · Answer

I'll try it again from scratch and see what happens

baldymcgee6 · Answer

I tried that as well... Wolfram's explanation:
y' = -x/y 
derivative of x =1
y'' = -1/y

baldymcgee6 · Answer

what if we went:
y' = -xy^-1

anonymous · Answer

Should be the same if you use product rule.

baldymcgee6 · Answer

Yeah, i'm sure that the original y'' I found was correct, but it just bugs me how they simplified it so nicely.

anonymous · Answer

Pff, yeah, beats me.

baldymcgee6 · Answer

Yeah, thanks for the help, i'll leave it up to see if anyone else has any bright ideas.

baldymcgee6 · Answer

@satellite73, @Hero, @AccessDenied, @hartnn, @lgbasallote

hartnn · Answer

wolfram is showing me correctly as -x/y

anonymous · Answer

What about second derivative, @hartnn ?

baldymcgee6 · Answer

@hartnn the first derivative of x^2 + y^2 =1 is -x/y, but wolfram somehow simplified the second derivative into -1/y