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Implicit differentiation sin(x+y)=xy
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Im stuck where I have \[(1+\frac{ dy }{ dx })-x \frac{ dy }{ dx }=y+x \frac{ dy }{ dx }\]
sin(x+y)=xy or, cos(x+y) . [ 1+ dy/dx ] = x dy/dx + y
Ya thats what I ment
I dont know what to do from here exactly in cos(x+y) [1+dy/dx]= x dy/dx + y
or, cos(x+y) + cos(x+y) dy/dx = x dy/dx + y or, cos(x+y) dy/dx - x dy/dx = y - cos(x+y) or, dy/dx {cos(x+y) - x} = y - cos(x+y) or, dy/dx = [ y - cos(x+y) ] / [ cos(x+y) - x ].................. Ans.
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The last one
differentiation of sin(x+y) is :- cos(x+y) . [ 1+ dy/dx ]
right
Yes the Last one [ y - cos(x+y) ] / [ cos(x+y) - x ] is the answer :)
Thanks for the help :)
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