Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

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Homework Help Overview

The discussion revolves around finding dy/dx using implicit differentiation for two equations: 6x² + 8xy + y² = 6 and 3x² = (2 - y) / (2 + y). Participants are exploring the application of differentiation rules, including the product rule and quotient rule, while addressing potential errors in their calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to differentiate the given equations, applying rules such as the product and quotient rules. There are questions about the correct application of the chain rule and the handling of terms during differentiation. Some participants express uncertainty about specific steps and calculations, while others point out potential errors in reasoning.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the proper application of differentiation rules, and there is an exploration of different interpretations of the derivative calculations. However, there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of homework assignments, which may limit the information they can use or the methods they can apply. There is also a focus on checking answers against provided solutions, which may influence their reasoning processes.

staples82
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Homework Statement


Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6


Homework Equations


n/a


The Attempt at a Solution


I'm using y'=(dy/dx)

I found the derivative of the above problem.
12x+8xy'+10y=0 (I used the product rule to find the derivative of 8xy')

12x+8xy'+10y=0

8xy'=-12x-10y

xy'=(-12x-10y)/(8x)

I checked the answer in the back of the book and it was y'=(-6x-4y)/(4x+y)

I'm not sure where I missed a step or a calculation error

Thanks!
 
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You forgot the chain rule on your y^2 term.
 
6x^2+8xy+y^2=6

12x+8xy' +8y + 2y**=0

This is wrong ...
 
i see what you tried to do. u tried to combine the 8y and 2y(dy/dx). they cannot be added together.

p.s.
next time u get stuck on something like this u can always work backwards. you could take the book's answer and multiply the top and bottom by 2. then u can figure it out from there
 
First of, thanks for all the help on my previous question, I have another one!

1. Homework Statement
Find dy/dx by implicit differentiation: 3x^2=(2-y)/(2+y)


2. Homework Equations
n/a


3. The Attempt at a Solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
First I used the quotient rule...
6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2

6x=(2y'+yy'-2y'+yy')

6x=2yy'/(2+yy')^2

After this, I got into a jumbled mess, I wasn't sure if I had to add y' to the bottom of the quotient rule.


I checked the answer in the back of the book and it was y'=-3x(2+y)^2/2


Thanks!
 
staples82 said:
...
6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2

Don't forget that the first term in the numerator involves the derivative of (2 - y), so the quotient should be

[(2+y)(-y')-(2-y)(y')]/(2+y)^2 .

You could multiply by (2+y)^2 to get

6x · [(2+y)^2] = -2y' - yy' - 2y' + yy' = -4y'

and solve for y' .
 
staples82 said:
First of, thanks for all the help on my previous question, I have another one!

1. Homework Statement
Find dy/dx by implicit differentiation: 3x^2=(2-y)/(2+y)


2. Homework Equations
n/a


3. The Attempt at a Solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
First I used the quotient rule...
6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2
Sign error: what's the derivative of (2-y) ?

6x=(2y'+yy'-2y'+yy')
What happened to the denominator on the RHS?

6x=2yy'/(2+yy')^2
How did (2+y)^2 from before become (2+yy')^2 ?
 

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