Implicit Diff Example - # 4

[Jason76]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 4x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - 6xy^{2} + [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 4x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -4x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -4x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

 

[stapel]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 4x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - [(6xy^{2}) + 3x^{2}2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - (6xy^{2}) + y'[3x^{2}2y) ] + 4y^{3}y] = 0\)

\(\displaystyle y'[3x^{2}2y) ] + 4y^{3}y] = -4x^{3} + 6xy^{2})\)

\(\displaystyle y' = \dfrac{ -4x^{3} + 6xy^{2}}{3x^{2}2y}\)

What were the instructions? What are you doing? What is your question?

Seriously, we really can't read your mind. :shock:

 

[Jason76]

Is the answer right? Any errors?

 

[Jason76]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 4x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - 6xy^{2} + [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 4x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -4x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -4x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

This answer isn't the same as in the book: \(\displaystyle \dfrac{3xy^{2} - 4x^{3}}{2y^{3} - 3x^{2}y}\)

 

[Deleted member 4993]

Is the answer right? Any errors?

Answer to "what"? - you did not post the whole question!!

 

[stapel]

Is the answer right?

We can't know, until you tell us what the question was. What were the instructions? What was the original exercise statement? What are you supposed to be doing with what original expression or conditions?

Seriously, we really can't read your mind. :shock:

 

[srmichael]

Is the answer right? Any errors?

Yes. Second line. Derivative of very first term. Derivative of 2x^4 is not 4x^3.

I did not bother to check the rest of your derivation once I saw that error.

 

[Deleted member 4993]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 4x^{3}\)\(\displaystyle - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\) .... Incorrect

\(\displaystyle 4x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 4x^{3} - 6xy^{2}\) + \(\displaystyle [3x^{2}2yy' \)+ 4y^{3}y'] = 0[/tex].... Incorrect sign

\(\displaystyle 4x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -4x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -4x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

.

 

[Jason76]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) - (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} - 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle -y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle -y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

\(\displaystyle (-1)-y' = (\dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}})(-1)\) How to get rid of that negative sign on the left.

 

[srmichael]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) - (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} - 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle -y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle -y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

\(\displaystyle (-1)-y' = (\dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}})(-1)\) How to get rid of that negative sign on the left.

How come I always feel like I have run a marathon when I look at one of your threads?

Anyway, your second line is incorrect. It should be:

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

But then you remarkabley make an arithmetic error later that negates the error in the second line. Awesome.

To answer you question, divide by -1. How else? Ok, multiple by -1. Same diff.

 

[Deleted member 4993]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) - (3x^{2})(2yy') ] + 4y^{3}y' = 0\) ................ Incorrect - does not follow from line above

\(\displaystyle 8x^{3} - [6xy^{2} - 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - [3x^{2}2yy' + 4y^{3}y'] = 0\)................ Incorrect- does not follow from line above

\(\displaystyle 8x^{3} - 6xy^{2} - y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle -y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle -y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

\(\displaystyle (-1)-y' = (\dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}})(-1)\) How to get rid of that negative sign on the left.

.

 

[Jason76]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} + [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\) - This still isn't the answer of: \(\displaystyle \dfrac{3xy^{2} - 4x^{3}}{2y^{3} - 3x^{2}y}\)

 

[srmichael]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} + [3x^{2}2yy' + 4y^{3}y'] = 0\) ====> WRONG!

\(\displaystyle 8x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\) - This still isn't the answer of:

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - 3x^{2}2yy' + 4y^{3}y' = 0\)

Now continue...

 

[Jason76]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle -y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle -y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

\(\displaystyle (-1)-y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}(-1)\)

 

[JeffM]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} + [3x^{2}2yy' + 4y^{3}y'] = 0\) This line does not follow from the line above.

\(\displaystyle 8x^{3} - 6xy^{2} + y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\) - This still isn't the answer of: \(\displaystyle \dfrac{3xy^{2} - 4x^{3}}{2y^{3} - 3x^{2}y}\)

This is JUST algebra.

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0 \implies\)

\(\displaystyle 8x^{3} - \{(y^{2})(6x) + (3x^{2})(2yy')\} + 4y^{3}y' = 0 \implies\)

\(\displaystyle 8x^{3} - 6xy^{2} - 6x^{2}yy' + 4y^{3}y' = 0 \implies\)

\(\displaystyle 8x^{3} - 6xy^{2} = 6x^{2}yy' - 4y^{3}y' \implies\)

\(\displaystyle y' = \dfrac{8x^{3} - 6xy^{2}}{6x^2y - 4y^3} = \dfrac{-2(3xy^2 - 4x^3)}{-2(2y^3 - 3x^2y)} =\dfrac{3xy^2 - 4x^3}{2y^3 - 3x^2y} .\)

 

[srmichael]

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - [6xy^{2} + 3x^{2}2yy' ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - [3x^{2}2yy' + 4y^{3}y'] = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - y'[3x^{2}2y + 4y^{3}] = 0\)

\(\displaystyle -y'[3x^{2}2y + 4y^{3}] = -8x^{3} + 6xy^{2}\)

\(\displaystyle -y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}\)

\(\displaystyle (-1)-y' = \dfrac{ -8x^{3} + 6xy^{2}}{3x^{2}2y + 4y^{3}}(-1)\)

OH FOR HEAVEN'S SAKES!!

\(\displaystyle 2x^{4} - 3x^{2}y^{2} + y^{4} = 0\)

\(\displaystyle 8x^{3} - [(y^{2})(6x) + (3x^{2})(2yy') ] + 4y^{3}y' = 0\)

\(\displaystyle 8x^{3} - 6xy^{2} - 6x^{2}yy' + 4y^{3}y' = 0\)

\(\displaystyle 4y^{3}y' - 6x^{2}yy' = 6xy^{2} - 8x^{3}\)

\(\displaystyle y'(4y^{3} - 6x^{2}y) = 6xy^{2} - 8x^{3}\)

\(\displaystyle y' = \dfrac{6xy^{2} - 8x^{3}}{4y^{3} - 6x^{2}y}\)

\(\displaystyle y' = \dfrac{3xy^{2} - 4x^{3}}{2y^{3} - 3x^{2}y}\)

 

[Jason76]

So the goal is to seperate the y' numbers from the others, and place them on one side of the equation by themselves. After that, isolate y prime and work from there.

 

[lookagain]

Jason76, please stop your habit of not posting complete problems in your original posts. You ignored the comments of others in post numbers 2, 5, and 6 in this thread and didn't post the full instructions as part of the problems. For example, in your original post, you could have included, "Solve for y' using implicit differentiation." You and the other users can then make better use of your time and our time in helping you understand your problem-solving techniques.