Double derivate implisitt function

[Phat]

We got:

solve by x.

y^3+x^2y^2-y*sinx=1

derivate:

3y^2*y' + 2xy^2 + 2yx^2*y' - (y'sinx + ycosx) = 0


but to derivate this once again I am not sure how to proceed.

My solution:

[3y^2*y'+6y*y''] + [2y^2 + 4xyy'] + ...

solutoin manual:

6y*y'' + 2x2yy' + [2x^2*1*y'+2x^2*y*y'']



I've been searching for websites where I can read about this, but no luck.

What I think is done in the solution manual is that when x is x^1 it cannot be further derivated, because we kinda "loose" information?

I need some guidance or explanation on how to handle equations that got:

y*y'

x*y^2

x*y*y'


Thanks in advance!

 

[HallsofIvy]

We got:

solve by x.

y^3+x^2y^2-y*sinx=1

derivate:

3y^2*y' + 2xy^2 + 2yx^2*y' - (y'sinx + ycosx) = 0

Okay, good!


[quiote]but to derivate this once again I am not sure how to proceed.

My solution:

[3y^2*y'+6y*y''] + [2y^2 + 4xyy'] + ...[/quote]
You have the y' and y" reversed in the first braces
It is (3y^2)(y')'+ (3y^2)'(y')= (3y^2)y"+ 6yy'.

For the second (2xy^2)'= (2x)'(y^2)+ (2x)(y^2)'= 2y^2+ (2x)(2yy')= 2y^2+ 4xyy'.

For the third term (2yx^2y')'= (2y)'(x^2)(y')+ (2y)(x)'(y')+ (2y)(x)(y')'= 2x^2y'+ 2yy'+ 2yxy".

For the fourth term (-y'sin x)'= (-y')'(sin x)+ (-y)(sin x)'= -y"sin x- y cos x.

For the fourth term (-ycos x)'= (-y)'(cos x)+ (-y)(cos x)'= -y' cos x+ y sin x.

Combine those.

solutoin manual:

6y*y'' + 2x2yy' + [2x^2*1*y'+2x^2*y*y'']



I've been searching for websites where I can read about this, but no luck.

What I think is done in the solution manual is that when x is x^1 it cannot be further derivated, because we kinda "loose" information?

I need some guidance or explanation on how to handle equations that got:

y*y'

x*y^2

x*y*y'


Thanks in advance!

 

[Phat]

Thanks! That made a lot more sense than the teachers solution. I get it 100% now =)