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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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