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xcosy=x^2+y^3 find y'
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You have some chain rule in your near future. Let's see your best efforts.
\[x(-siny)y' + cosyy = 2x + 3y^2y'\]
That's awesome! Good work. Now some algebra and you'll be done.
then i simplified it, i got \[y'=(2x-cosy)/(-3y^2-xsiny)\]
You could pretty that up a bit.
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It's a multiple choice worksheet, that's the answer form the have it in, but the only difference is that the -3y^2 is positive is that a typo on the worksheet?
Nah, that's silly. I would go with \(y' = \dfrac{\cos(y) - 2x}{3y^{2}+x\sin(y)}\), if I had to go with something.
yeah but that isn't there either O.o
Can't help that. You got it! Score: You: 1 Sheet: 0
Thanks!! Appreciate the help
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