Implicit Differentiation - Stuck on simplifying

kakariki

New member
Joined
Jun 14, 2010
Messages
3
Hello,

Thank you for taking the time to read my post. I am following along with some example problems in my textbook, and I am very confused why I am getting a different answer than the textbook.

Problem is to find y'.
[MATH] x^3 + y^3 = 6xy[/MATH].

Solution:
Going to skip a bit on the solution to where I have the issue. Brief explanation - use sum rule and chain rule on left side to get:
[MATH] 3x^2 + 3y^2 y' [/MATH].
Right side use product rule to get:
[MATH] 6(xy' + y) [/MATH]Doing a bit of simplification we end up with:
[MATH] x^2 + y^2y' = 2xy' + 2y [/MATH]
The reason I am skipping the solution to this part is because this is essentially where I start to deviate from the textbook. I decided to solve for y' myself, without looking at what the textbook did, and I am very confused as to why I am not getting the same answer.

So, below is my attempt to solve for y'. I believe I am getting the wrong answer (please confirm) and I cannot figure out why.

Solving for y':
[MATH] x^2 + (y^2 y') = 2xy' + 2y [/MATH]

Step 1 - Move like terms onto the same side of the equal sign:
[MATH]x^2 - 2y = (2xy') - (y^2 y') [/MATH]
Step 2 - Factor out y' on the right hand side:
[MATH] x^2 - 2y = y'(2x - y^2) [/MATH]
Step 3 - Divide both sides by 2x-y^2:
[MATH] (x^2 - 2y)/(2x - y^2) = y' [/MATH]

When I looked up from this to check my work I saw that the textbook had done the following:
Solving from [MATH] x^2 + (y^2 y') = (2x y' + 2y) [/MATH]
1. Move like terms onto same side of the equal sign. Note that opposite terms were moved from how I did it.
[MATH] y^2 y' - 2xy' = 2y - x^2 [/MATH]
2. Factor y' out of the left hand side:
[MATH] y'(y^2 - 2x) = 2y - x^2 [/MATH]
3. Divide both sides by y^2 - 2x:
[MATH] y' = (2y - x^2) / (y^2 - 2x) [/MATH]

I don't understand what I did wrong and I'm feeling a bit like an idiot for not seeing it. I'm hoping you can help me discover my error.
 
Hello,

Thank you for taking the time to read my post. I am following along with some example problems in my textbook, and I am very confused why I am getting a different answer than the textbook.

Problem is to find y'.
[MATH] x^3 + y^3 = 6xy[/MATH].

Solution:
Going to skip a bit on the solution to where I have the issue. Brief explanation - use sum rule and chain rule on left side to get:
[MATH] 3x^2 + 3y^2 y' [/MATH].
Right side use product rule to get:
[MATH] 6(xy' + y) [/MATH]Doing a bit of simplification we end up with:
[MATH] x^2 + y^2y' = 2xy' + 2y [/MATH]
The reason I am skipping the solution to this part is because this is essentially where I start to deviate from the textbook. I decided to solve for y' myself, without looking at what the textbook did, and I am very confused as to why I am not getting the same answer.

So, below is my attempt to solve for y'. I believe I am getting the wrong answer (please confirm) and I cannot figure out why.

Solving for y':
[MATH] x^2 + (y^2 y') = 2xy' + 2y [/MATH]
Step 1 - Move like terms onto the same side of the equal sign:
[MATH]x^2 - 2y = (2xy') - (y^2 y') [/MATH]
Step 2 - Factor out y' on the right hand side:
[MATH] x^2 - 2y = y'(2x - y^2) [/MATH]
Step 3 - Divide both sides by 2x-y^2:
[MATH] (x^2 - 2y)/(2x - y^2) = y' [/MATH]

When I looked up from this to check my work I saw that the textbook had done the following:
Solving from [MATH] x^2 + (y^2 y') = (2x y' + 2y) [/MATH]
1. Move like terms onto same side of the equal sign. Note that opposite terms were moved from how I did it.
[MATH] y^2 y' - 2xy' = 2y - x^2 [/MATH]
2. Factor y' out of the left hand side:
[MATH] y'(y^2 - 2x) = 2y - x^2 [/MATH]
3. Divide both sides by y^2 - 2x:
[MATH] y' = (2y - x^2) / (y^2 - 2x) [/MATH]

I don't understand what I did wrong and I'm feeling a bit like an idiot for not seeing it. I'm hoping you can help me discover my error.
Your answer and the book's answer are EXACTLY equivalent.

If you multiply your solution by \(\displaystyle \frac{-1}{-1}\) which is equivalent of multiplying by (1) - you'll get book's solution.
 
Your answer and the book's answer are EXACTLY equivalent.

If you multiply your solution by \(\displaystyle \frac{-1}{-1}\) which is equivalent of multiplying by (1) - you'll get book's solution.

Thank you. I figured it would be something like that. I really appreciate the help.
 
Top