# implicit differentiation

• Mar 10th 2011, 11:01 AM
Foxlion
implicit differentiation
Problem: $x^2+xy-y^2=4$

Solution attempt: $\frac{d}{dx} (x^2)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$

$2x+y+xy'-2yy'=0$

$(2x+y)+(xy'-2yy')=0$

$y'(x-2y)=-(2x+y)$

$y'=\frac{-(2x+y)}{x-2y}$

$y'=\frac{-2x-y}{-2y+x}$

$y'=\frac{2x-y}{2y+x}$

Answer in book: $y'=\frac{2x+y}{2y-x}$

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?
• Mar 10th 2011, 11:36 AM
icemanfan
Your second to last step is correct. What you are doing when you "cancel out" the negatives is multiplying the fraction by $\frac{-1}{-1}$. So you get $\frac{-1}{-1} \cdot \frac{-2x - y}{-2y + x} = \frac{(-1)(-2x - y)}{(-1)(-2y + x)}$. Then use the distributive property to arrive at the correct answer (the one in your book).
• Mar 10th 2011, 11:39 AM
profound
$\frac{d}{dx} (x^2)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$

$2x+y+xy'-2yy'=0$

$(2x+y)=2yy'-xy'$

$(2x+y)=y'(2y-x)$

$y'=\frac{2x+y}{2y-x}$

which matches your answer: $y'=\frac{2x+y}{2y-x}$
• Mar 10th 2011, 11:45 AM
Foxlion
case closed, thank you both