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Mathematics 11 Online
OpenStudy (anonymous):

xcosy=x^2+y^3 find y'

OpenStudy (tkhunny):

You have some chain rule in your near future. Let's see your best efforts.

OpenStudy (anonymous):

\[x(-siny)y' + cosyy = 2x + 3y^2y'\]

OpenStudy (tkhunny):

That's awesome! Good work. Now some algebra and you'll be done.

OpenStudy (anonymous):

then i simplified it, i got \[y'=(2x-cosy)/(-3y^2-xsiny)\]

OpenStudy (tkhunny):

You could pretty that up a bit.

OpenStudy (anonymous):

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?

OpenStudy (tkhunny):

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.

OpenStudy (anonymous):

yeah but that isn't there either O.o

OpenStudy (tkhunny):

Can't help that. You got it! Score: You: 1 Sheet: 0

OpenStudy (anonymous):

Thanks!! Appreciate the help

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