Taking derivatives of DE to find higher derivatives

[shemer77]

Homework Statement

http://gyazo.com/d875970963124b7d4a64acc887f168fa

The Attempt at a Solution

I don't 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''

 

[tiny-tim]

hi shemer77!

(try using the X² button just above the Reply box )

y'''=(y'')'=(1-y²)'=-2yy'(this part)

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)'

y⁽⁴⁾=(y''')'=(-2yy')'=-2y'y'-2yy"(huh?)=-2(y')^2-2yy''

similarly, use the product rule for (-2yy')'

 

[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?

 

[tiny-tim]

… for the first one what I've been taught is
(1-y²)'=1'-y²'=-2y
where does that extra y' come from?

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' ​