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How would I solve for $y'$ using implicit differentiation?

$x^2 + 2xy -y^2 + x = 2$

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2 Answers 2

up vote 2 down vote accepted

Product rule, chain rule and power rule will get you the expression in terms of $x,y,$ and $y'$. Gather all your $y'$ terms on one side, factor it out, then divide by the other factor to solve for $y'$.

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Can you please write out all of the steps? – user44816 Nov 12 '12 at 1:09
Do you know the rules I mentioned? – Cameron Buie Nov 12 '12 at 1:12
OP, can you write out any of the steps?? – The Chaz 2.0 Nov 12 '12 at 1:14
Yes, I tried solving like this: 2x + (2x*y' + 2*y) - 2y *y' + 1 = 0. Then I got stuck. What would be the next step? – user44816 Nov 12 '12 at 1:25
Gather all of the $y'$ terms on one side, factor out the $y'$, and divide both sides to get the $y'$ alone. Does that help? Your work looks correct so far. – Todd Wilcox Nov 12 '12 at 1:37

Differentiate both sides with respect to $x$, keeping the chain rule in mind, to get $$2x+2y+2xy'-2yy'+1=0$$

Now isolate $y'$ and we're done: $y'(2x-2y)=-2x-2y-1$. Therefore, $y'=?$

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