Taking derivatives of DE to find higher derivatives

1. Jul 6, 2012

shemer77

1. The problem statement, all variables and given/known data
http://gyazo.com/d875970963124b7d4a64acc887f168fa

3. The attempt at a solution
I dont understand the part where they take the derivative of a diff eq to find the higher derivatives can somebody explain it?
Mainly the answer where they get
y''=(y'')'=(1-y^2)'=-2yy'(this part)
y^(4)=(y''')'=(-2yy')'=-2y'y'-2yy"(huh?)=-2(y')^2-2yy''

2. Jul 6, 2012

tiny-tim

hi shemer77!

(try using the X2 button just above the Reply box )
since you have an equation for y'', you use it to simplify all the higher derivatives

(1-y2)' = (-y2)' = (use the chain rule) (-2y)(y)'
similarly, use the product rule for (-2yy')'

3. Jul 6, 2012

shemer77

Hmm ok I think I understand the second one but for the first one what I've been taught is
(1-y2)'=1'-y2'=-2y
where does that extra y' come from?

4. Jul 6, 2012

tiny-tim

remember that ' means d/dx, not d/dy

so the chain rule gives you

y2' = d/dx (y2)

= d/dy (y2) * dy/dx

= 2y * y'