# Implicit differentiation problem.

1. Apr 28, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

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

x2-√(xy)+y2=6

2. Relevant equations

3. 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 seperate 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.

2. Apr 28, 2012

### Dick

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

3. Apr 28, 2012

### scurty

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.

4. Apr 28, 2012

### charmedbeauty

ok yeah the 6 will cancel.

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

"" " " "" = y-2x

??

5. Apr 28, 2012

### scurty

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.

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.