Finding the derivative dy/dx for y^2 = 2 + xy

[Math wiz ya rite 09]

Consider the curve given by y^2 = 2 + xy.

Find dy/dx.

I remember that like when you take the derivative of a y that you must multiply it by a dy/dx?

Could someone help me out? Thanks

 

[arthur ohlsten]

y^2=2+xy rewrite
y^2-xy=2 complete the square
y^2-xy+[x/2]^2=2+[x/2]^2
[y-x/2]^2=2+x^2/4
[y-x/2]^2=[8+x^2]/4 take square root
y-x/2=1/2[x^2+8]^1/2
y=x/2+[x^2+8]^1/2 /2
2y=x+[x^2+8]^1/2 take derivative
2 dy/dx= 1+1/2[x^2+8]^-1/2 [2x]
2dy/dx = 1 +x/[x^2+8]^1/2
dy/dx=1/2 + x/[2[x^2+8]^1/2 ] answer
Arthur

 

[pka]

\(\displaystyle \L \begin{array}{l}
2yy' = y + xy' \\
y' = \frac{y}{{2y - x}} \\
\end{array}\)

Would is not be nice if those who would try to help could at least learn LaTex.
Trying to read Authur’s post gives me a headache.

 

[Math wiz ya rite 09]

pka. what kind of method did you use?

 

[Math wiz ya rite 09]

im getting the same results as pka, but some work would be useful.

 

[Math wiz ya rite 09]

i figured it out!!

 

[arthur ohlsten]

I am sorry you are getting a headache.

I completed the square so as to have a equation of y as a function of x.
THIS IS THE SAME AS THE QUADRATIC EQUATION

y^2=2+xy
y^2-yx -2=0
by the quadratic equation:
a=1
b=-x
c=-2
then:
y=-b+/-[b^2-4ac]^1/2 all over 2

y=x+/- sqrt[x^2+8] all over 2
2y=x+/-[x^2+8]^1/2 take derivative
2 dy/dx = 1+/- 1/2[x^2+8]^-1/2 [2x]
2 dy/dx = 1+/- x/[x^2+8] answer

The other answers you received contained y terms in the derivative.
I doubt if your instructor will accept them as correct answers .
Arthur

 

[arthur ohlsten]

Perhaps your instructor should have told you he expected y' as a function of x, and not as a function of x and y.

Solve for y as a function of x, before obtaining the derivative.
You can solve for y[x] by the quadratic equation, or by completing the square.

Completeing the square is a good technique to become familiar with, because it will help you later on when dealing with conic sections.

Arthur