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

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

2 Answers 2

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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?2012-11-12
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    Do you know the rules I mentioned?2012-11-12
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    OP, can you write out **any** of the steps??2012-11-12
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    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?2012-11-12
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    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.2012-11-12
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    I got to this point: (2x*y' + 2*y)-2yy' = -2x-1. How would I continue factoring at this point to get the y' alone?2012-11-12
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    @user44816 You didn't try what Cameron and Todd said to try. Why don't you try that?2012-11-12
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    @user44816: You're very close! You want to gather all the $y'$ terms on one side, though. I see that you gathered all the terms with a $y$ **or** a $y'$ on one side, but we want **just** the $y'$ terms. One of the terms you've got on the left-hand side still needs to be moved to the other side--in particular, the one without a $y'$ in it.2012-11-12
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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, $y'=?$