Question

Can someone help me with the next step of this problem: I am doing implicit differentiation:

3(2)(x^2+y^2)(2x+2yy')=100(y+xy')

Answer 1

You just want to isolate y'?

Answer 2

yes

Answer 3

Okay. Let me type this out:
Start by foiling out the left side, distributing the 6, and distributing the 100 on the right.
\[12x^3+12y^3y'+12xy^2+12yx^2y'=100y+100xy'\]
Then get all the y' terms on one side.
\[12y^3y'+12yx^2y'-100xy'=100y-12x^3-12xy^2\]
Then factor out a y':
\[y'(12y^3+12yx^2-100x)=100y-12x^3-12xy^2\]
Divide over:
\[y'=\frac{100y-12x^3-12xy^2}{12y^3+12yx^2-100x}\]
Then you can re-plug in your original y to get it complete in terms of x.

Answer 4

Oh, so on the left side, we must multiply out (x^2+y^2)(2x+2yy')?

Answer 5

Yes

Answer 6

and then multiply the 6 in

Answer 7

Or you can divide the 6 over. But then you have to carry around a 50/3 :/

Answer 8

okay thanks

Answer 9

No problem :)