Math CentralQuandaries & Queries


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.


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