1. Apr 19, 2005

### 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

2. Apr 19, 2005

### whozum

I'll go through, cant guarantee accuracy though.

$$Third = xy^3 + 2$$

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

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

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

3. Apr 19, 2005

### whozum

4. Apr 19, 2005

### 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
$$(6y^2y')'=12yy'+6y^2y''$$
$$(6xyy'y')'=6yy'y'+6xy'y'y'+6xyy''y'+6xyy'y''$$
$$(3xy^2y'')'=3y^2y''+6xyy''y'+3xy^2y'''$$
$$12yy'+6y^2y''+6y(y')^2+6y(y')^3+18xyy'y''+3y^2y''+3xy^2y'''$$

5. Apr 19, 2005

### whozum

for $$6xy(y')^2$$

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

Then:

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

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

6. Apr 19, 2005

### krusty the clown

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