Question

How do you implicitly differentiate `-1=xy+cot^2(x-y)`?

Answer 1

`y'=(2cot(x-y)-ysin(x-y))/(xsin(x-y)+2cot(x-y))`

Explanation:

`0=y+xy'-2cot(x-y)/sin(x.y)(1-y')`

`0=y-2cot(x-y)/sin(x-y)+y'(x+2cot(x-y)/sin(x-y))`
Multiplying by

`sin(x-y)`

`2cot(x-y)-ysin(x-y)=y'(xsin(x-y)+2cot(x-y))`

so we get

`y'=(2cos(x-y)-ysin(x-y))/(xsin(x-y)+2cot(x-y))`