What is the derivative of y=x+sin(xy) in terms of x and y at any point on the curve

So I got to y'\1+cos(xy)*....

I dont know what to do next

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asked by Anonymous
  1. Sorry that is suppose to say y'=

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    posted by Anonymous
  2. y=x+sin(xy)
    You will need the product rule for the sine term

    dy/dx = 1 + cos(xy)(xdy/dx + y)
    the second dy/dx is imbedded in a product so we have to expand
    I will use y' for dy/dx
    y' = 1 + xcos(xy)*y' + ycos(xy)
    y' - xcos(xy) y' = 1 + ycos(xy)
    y' (1 - xcos(xy) ) = 1 + ycos(xy)
    y' = (1 + ycos(xy))/(1 - xcos(xy) )


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    posted by Reiny

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