find dy/dx of sin(xy)=x+y So far I have cos(xy)*(xy'+1y)=1+y' (cos(xy))xy'+(cos(xy))y=1+y' Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y. Would it be y=((1+y')-(cos(xy)xy'))/(cos(xy)) ???
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Originally Posted by AlphaTauLambda find dy/dx of sin(xy)=x+y So far I have cos(xy)*(xy'+1y)=1+y' (cos(xy))xy'+(cos(xy))y=1+y' Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y. Would it be y=((1+y')-(cos(xy)xy'))/(cos(xy)) ??? solve for y', which is dy/dx ... not y
Skeeter. Thanks. I go to school in North Texas... haha The person who helped me with this problem has written (xcosxy-y)y'=1-ycosxy I don't understand the transition from the previous step (which is (cosxy)xy' + (cosxy)*y=1+y' Any ideas?
distribute the on the left ... get the terms w/ on the same side ... factor out ...
Thanks a million, Skeeter! This makes perfect sense!!!
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