Implicit Differentiation: Understanding Higher Order Derivatives

[krusty the clown]

y^{(3)}= 2+xy^3
y^{(4)}=y^3+3xy^2y'
y^{(5)}=6y^2y'+6xy(y')^2+3xy^2y''
y^{(6)}=18y(y')^2+6y^2y''+6x(y')^3+12xyy'y''

It has been awhile since I have done implicit differentiation, and I am not quite sure if I have used the chain rule properly in each step. I would greatly appreciated any help you could give me on this.

-Erik

 

[whozum]

I'll go through, can't guarantee accuracy though.

Third = xy^3 + 2

Fourth = 3y^2xy' + y^3

Fifth = 3y^2y' + 3((2xyy'+y^2)y' + y''(y^2x))

Sixth = 6y(y')^2+y''(3y^2) + 2xyy'y''+y'''(y^2x) + 3((y''(2xyy'+y^2) + y'(y''(2xy)+2yy')+2yy')

 

[whozum]

I checked the first two on http://www.calc101.com/webMathematica/derivatives.jsp and they came out good, that last one is a mess but you can check it yourself.

 

[krusty the clown]

thanks, I made a mistake somewhere in the sixth but when I went through it again I still get a different answer than yours. Does that link do implicit differentiation, I didn't see it anywhere. Anyway, it isn't that important. Again, thanks for your help.

Dang, now I am currious...


y^{(5)}=6y^2y'+6xy(y')^2+3xy^2y''
so for the individual terms we should get
(6y^2y')'=12yy'+6y^2y''
(6xyy'y')'=6yy'y'+6xy'y'y'+6xyy''y'+6xyy'y''
(3xy^2y'')'=3y^2y''+6xyy''y'+3xy^2y'''
added together, I get
12yy'+6y^2y''+6y(y')^2+6y(y')^3+18xyy'y''+3y^2y''+3xy^2y'''

 

[whozum]

for 6xy(y')^2

I set u = 6xy, v = (y')^2

Then:

d(uv)/dx = v du/dx + u dv/dx

6(y')^2(y+xy') + 6xy(2y'y'')

 

[krusty the clown]

If your answer is expanded and mine is compressed they are the same for that term.