Expert: Scotto - 10/29/2009

Question
QUESTION: Hello - I am a high school cal. teacher and I started working through this problem and then started second guessing myself.  Basically I'd like someone to look this over and tell me if I've made any mistakes throughout.  Thanks.

The problem states:  Find y" if 3x^2-xy^2 + y^-1 = 1

I started with:
6x - x(2yy') + (y^2)(-1) - (1)(y^-2)(y') = 0

I continued on this path to find y'

I then got y' = (y^4 - 6xy^2)/ (-2xy^3 - 1)

How am I doing so far??

Anyway - so I then continue from there and get:

y" = (-2xy^3 - 1)(4y^3y' - 6x(2yy')+y^2(-6)) - (y^4 - 6xy^2)[(-2x)(3y^2y') + y^3(-2)] all divided by (-2xy^3 - 1)^2

Of course I kept working, simplifying etc and then substituted my original y' into the equation.

so the final answer that I got was:
y" = (-6xy^10 - 48x^2y^8 - 6y^7 + 72x^3y^6 - 72x^2y^3 - 6y^2) / (-2xy^3 - 1)^3

Am I even in the ball park?  Thanks!

ANSWER: You're right at this point: 6x - x(2yy') + (y^2)(-1) - (1)(y^-2)(y') = 0
I believe we need to show our work more.  That almost sounds like a teacher's line...

Factoring out y' and subtracting off the rest from both sides gives y'(-2xy - y^-2) = -6x - y^-2.

Multiplying by -y² gives y'(2xy³ + 1) = (6xy² + 1).

Dividing both sides by (2xy³ + 1) gives y' = (6xy² + 1)/(2xy³ + 1).

Now to take the second derivative, which is a quotient rule with a couple of product rules nested inside.  
That is, y" = ((2xy³ + 1)((6x)2yy' + 6y²) - (6xy² + 1)(6xy²y' + 2y³))/(2xy³ + 1)².

Dropping the () around the 6x gives
y" = ((2xy³ + 1)(12xyy' + 6y²) - (6xy² + 1)(6xy²y' + 2y³))/(2xy³ + 1)².

Multiplying it out the top gives
(24x²y^4•y' + 12xyy' + 12xy^5 + 6y²) - (36x²y^4•y' + 6xy²y' + 12xy^5 + 2y³).
The 1st term in each combines.  The 3rd term  in each cancel.

That makes this combines to (-12x²y^4•y' + 12xyy' + 6y² - 6xy²y' + 2y³).
Factoring out the y' gives ((-12x²y^4 + 12xy - 6xy²)y' + 6y² + 2y³).

We need to put (6xy² + 1)/(2xy³ + 1) in for y'.  This is messy, no?
That gives ((-12x²y^4 + 12xy - 6xy²)(6xy² + 1)/(2xy³ + 1) + 6y² + 2y³).

Remembering this was the numerator, I'll put the (2xy³ + 1) in the denominator and multiply
the terms at the end { 6y² + 2y³ } by (2xy³ + 1) .

This gives  ((-12x²y^4 + 12xy - 6xy²)(6xy² + 1) + (6y² + 2y³)(2xy³ + 1))/(2xy³ + 1)².

Now to multiply the top.  The 1st two terms gives
-72x³y^6 - 12x²y^4 + 72x²y³ + 12xy - 36x²y^4 - 6xy².

The 2nd two terms give 12xy^5 + 4xy^6 + 6y² + 2y³.

Combine to get -72x³y^6 - 12x²y^4 + 72x²y³ + 12xy - 36x²y^4 - 6xy² + 12xy^5 + 4xy^6 + 6y² + 2y³
Now recall that the denominator was (2xy³ + 1)².

I would then divide the numerator up by powers of y.
-72x³y^6 + 4xy^6 + 12xy^5 - 12x²y^4 - 36x²y^4 + 72x²y³ + 2y³ - 6xy² + 6y² + 12xy.

This done, factor out y to a power.  That is,
(-72x³+4x)y^6 + 12xy^5 - (12x²+36x²)y^4 + (72x²+2)y³ - (6x-6)y² + 12xy.

Once again, our denominator is (2xy³ + 1)².

I have to print this out and check it myself.


---------- FOLLOW-UP ----------

QUESTION: First of all - super thanks for all of the hard work. So here is my question... where in the following steps am I going wrong. (Btw - after checking the text again they only asked for the 1st derivative - THANK GOODNESS!!) :) - However for my own sanity I still must venture on to discover the second derivative and gain an understanding of the correct solution.
I go from :
6x - x(2yy') + (y^2)(-1) - (1)(y^-2)(y') = 0
to

6x - 2xyy' - y^2 - y'y^-2 = 0

I multiplied through by y^2 and got:

6xy^2 - 2xy^3y' - y^4 - y' = 0

which simplified to:

-2xy^3y' - y' = -6xy^2 + y^4

now I factored y' and got

y' (-2xy^3 - 1) = -6xy^2 + y^4

which leads to :

y' = (-6xy^2 + y^4) / (-2xy^3 - 1)  

Answer
Which is the same thing that I got, only the (-6xy^2 + y^4) was switched around.
I have this habit of putting positive numbers in front.

If you read through the rest of my answer, I believe I did y" correctly with the quotient rule.
This also involved using the chain rule in the middle.