Implicit differentiation problem.

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Homework Help Overview

The problem involves finding dy/dx in terms of x and y from the equation x² - √(xy) + y² = 6. This falls under the topic of implicit differentiation in calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the steps taken in differentiating the left-hand side of the equation and express uncertainty about the simplification of terms. Questions arise regarding the treatment of the constant on the right side of the equation and the proper handling of terms involving dy/dx.

Discussion Status

There is ongoing exploration of the differentiation process, with participants providing feedback on each other's attempts. Some participants suggest corrections to the approach, particularly regarding the treatment of constants and the organization of terms involving dy/dx.

Contextual Notes

Participants note potential errors in the original poster's differentiation steps and question assumptions about the simplification of terms. The discussion reflects a collaborative effort to clarify the implicit differentiation process without reaching a final solution.

charmedbeauty
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Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

then I thought I would multiply through to get a common denominator..

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] / [2√(xy) =6 -2x

so multiply through by denominator to simplify and collect like terms

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] = (6-2x)(2√(xy))

so taking that -y over

[2y(dy/dx)(2√(xy))-(x(dy/dx))]= (6-2x)(2√(xy))+y

taking out dy/dx as a common factor

dy/dx[4y√(xy) - x] = (6-2x)(2√(xy))+y

so dy/dx = [(6-2x)(2√(xy))+y] / [4y√(xy) - x]

Is this right, because I checked it on wolfram and it had a different answer so I guess not, can someone please shed some light on where I went wrong?

Thanks.
 
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charmedbeauty said:

Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

then I thought I would multiply through to get a common denominator..

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] / [2√(xy) =6 -2x

so multiply through by denominator to simplify and collect like terms

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] = (6-2x)(2√(xy))

so taking that -y over

[2y(dy/dx)(2√(xy))-(x(dy/dx))]= (6-2x)(2√(xy))+y

taking out dy/dx as a common factor

dy/dx[4y√(xy) - x] = (6-2x)(2√(xy))+y

so dy/dx = [(6-2x)(2√(xy))+y] / [4y√(xy) - x]

Is this right, because I checked it on wolfram and it had a different answer so I guess not, can someone please shed some light on where I went wrong?

Thanks.

After the step labelled "Simplifying the last term" why to you have 6 on the right side?
 
charmedbeauty said:

Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

Left hand side is good, but I'm not so sure you want to simplify the last term. What is ##\frac{d}{dx} (6)##?

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

Every term on the left hand side isn't multiplied by ##\frac{dy}{dx}##, you need to move the other term over as well before you can factor ##\frac{dy}{dx}## out.

Try fixing those errors and continue.
 
scurty said:
Left hand side is good, but I'm not so sure you want to simplify the last term. What is ##\frac{d}{dx} (6)##?



Every term on the left hand side isn't multiplied by ##\frac{dy}{dx}##, you need to move the other term over as well before you can factor ##\frac{dy}{dx}## out.

Try fixing those errors and continue.



ok yeah the 6 will cancel.

and when I take the 2x over I also take the -y over

"" " " "" = y-2x

??
 
I'm not sure what you mean by the 6 canceling. The derivative of a constant is 0.

This just becomes an algebra exercise from here. The y isn't a term by itself, it is being multiplied by a number.

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

Don't simplify the last term, multiply the last term out and then gather all the terms with ##\frac{dy}{dx}## onto one side and all the other terms onto the other side.
 

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