implicit differentiation

**Foxlion** · March 10th 2011, 11:01 AM

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?

**icemanfan** · March 10th 2011, 11:36 AM

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).

**profound** · March 10th 2011, 11:39 AM

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

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

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

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

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

which matches your answer: $y'=\frac{2x+y}{2y-x}$

**Foxlion** · March 10th 2011, 11:45 AM

case closed, thank you both

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