
implicit differentiation
Problem: $\displaystyle x^2+xyy^2=4$
Solution attempt:$\displaystyle \frac{d}{dx} (x^2)+\frac{d}{dx}(xy)\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$
$\displaystyle 2x+y+xy'2yy'=0$
$\displaystyle (2x+y)+(xy'2yy')=0 $
$\displaystyle y'(x2y)=(2x+y)$
$\displaystyle y'=\frac{(2x+y)}{x2y}$
$\displaystyle y'=\frac{2xy}{2y+x}$
$\displaystyle y'=\frac{2xy}{2y+x}$
Answer in book: $\displaystyle y'=\frac{2x+y}{2yx}$
Notes:I've always had a mundane problem with negatives and fractions, when I cancel out the negs in the last step, must I also switch the signs of the proceeding variables as well?

Your second to last step is correct. What you are doing when you "cancel out" the negatives is multiplying the fraction by $\displaystyle \frac{1}{1}$. So you get $\displaystyle \frac{1}{1} \cdot \frac{2x  y}{2y + x} = \frac{(1)(2x  y)}{(1)(2y + x)}$. Then use the distributive property to arrive at the correct answer (the one in your book).

$\displaystyle \frac{d}{dx} (x^2)+\frac{d}{dx}(xy)\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$
$\displaystyle 2x+y+xy'2yy'=0$
$\displaystyle (2x+y)=2yy'xy'$
$\displaystyle (2x+y)=y'(2yx)$
$\displaystyle y'=\frac{2x+y}{2yx}$
which matches your answer: $\displaystyle y'=\frac{2x+y}{2yx}$

case closed, thank you both