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I can't find an example in my book so I am not sure how I am suppose to do this.

I am trying to find the derivative of $y$ for $y+x\cos(y) = x^2y$

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    use chain rule.2011-10-09
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    So the xcosy is 1(cosy)+x(-siny)?2011-10-09
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    Close. Don't say "is," because you're not saying they're equal. And you are differentiating with respect to $x$, so the derivative of $\cos y$ - using the chain rule - is $(-\sin y)y'$. And that's just differentiating the left-hand side; now try what's on the right.2011-10-09
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    use the chain and product rules.2011-10-09
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    I got $y\prime + cosy(y\prime) - 2xy\prime y$2011-10-09
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    product rule: $(x \cos(y))'=x'\cos(y)+x(\cos(y))'$2011-10-09
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    product rule: $(x^2y)'=(x^2)'y+x^2y'$2011-10-09
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    I feel like there is a 2x missing from that.2011-10-09
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    are you thinking about $(x^2)'=2x$?2011-10-09
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    Yes that is what I meant.2011-10-09
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    @Austin: That's fair, though the post by OP shows little to no research effort. A simple google search brings up a good amount of information on the process.2011-10-10
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    That isn't really constructive since I did say I looked but I couldn't find anything.2011-10-10

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When you do implicit differentiation problems, there are three important things to keep in mind.

  1. Whenever you have a mixture of $x$ and $y$ factors, you must use the product/quotient rule.
  2. Whenever you differentiate a term involving $y$, you must include a factor of $y^\prime$ (since we are differentiating with respect to $x$, not $y$).
  3. If you did not differentiate a factor of $y$, you do not include a factor of $y^\prime$.

Keeping these things in mind, I get $$ 1 \cdot y^\prime + (\cos(y) - x\sin(y)y^\prime) = (2xy + x^2\cdot 1 \cdot y^\prime), $$ where I've included parentheses to show where product rule is taking place.

Now, the whole point of this business was to get $y^\prime$ by itself. So, move everything having to do with $y^\prime$ to one side of the equation and all other terms to the other. $$ y^\prime - x\sin(y)y^\prime - x^2y^\prime = 2xy - \cos(y). $$ Factoring out the $y^\prime$ gives $$ y^\prime(1 - x\sin(y) - x^2) = 2xy - \cos(y). $$ Finally, dividing to isolate $y^\prime$ leaves us with $$ y^\prime = \frac{2xy - \cos(y)}{1 - x\sin(y) - x^2}. $$