# Taking derivatives of DE to find higher derivatives

## Homework Statement

http://gyazo.com/d875970963124b7d4a64acc887f168fa

## 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''

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
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')' 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
Science Advisor
Homework Helper
… 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' 