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 Question from Monica, a student: Find dy/dx in terms of x and y, if sin(xy)=(x^2)-y. I tried to solve the problem by doing... =cos(xy)*[x*y'+y*x'] = 2x - y* y' =cos(xy)*[x*y'y+y*1] =2x-y*y' =cos(xy)*[(dy/dx)*x+y] = 2x-y*y' y'*(x+y)cos(xy) = y'(y-2x) I am not sure if I simplified the problem correctly, or distributed correctly. Could you please assist me with this practice problem? Thank you.

Hi Monica,

When you differentiate y in this expression you should get y' not y*y'. Hence the first line should be

$\cos(xy) \times [x \times y' + y \times x'] = 2x - y'$

which becomes

$\cos(xy) \times [x \times y' + y \times 1] = 2x - y'$

Can you see how to complete it now? Write back if you need more help.

Penny

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