Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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 $y'$.

share|cite|improve this answer
    
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
2  
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
2  
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, $y'=?$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.