How do you implicitly differentiate $-1=x-ycot^2(x-y)$ ?

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Answer 1 · 2017-06-18T19:09:34

$(dy)/dx =(1+2ycot(x - y) csc^2(x - y))/(cot^2(x-y)+2ycot(x - y) csc^2(x - y))$

Take the derivative, $d/dx$ of both sides

$d/dx (-1)=d/dx (x)-d/dx(ycot^2(x-y))$

$0=1-yd/dx(cot^2(x-y))-(dy)/dx cot^2(x-y)$

$(dy)/dx cot^2(x-y)+yd/dx(cot^2(x-y))=1$

$(dy)/dx cot^2(x-y)+y(-2 cot(x - y) csc^2(x - y))(1-dy/dx)=1$

$(dy)/dx cot^2(x-y)-2ycot(x - y) csc^2(x - y)+2ycot(x - y) csc^2(x - y)dy/dx=1$

$(dy)/dx cot^2(x-y)+dy/dx2ycot(x - y) csc^2(x - y)-2ycot(x - y) csc^2(x - y)=1$

$(dy)/dx (cot^2(x-y)+2ycot(x - y) csc^2(x - y))-2ycot(x - y) csc^2(x - y)=1$

$(dy)/dx (cot^2(x-y)+2ycot(x - y) csc^2(x - y))=1+2ycot(x - y) csc^2(x - y)$

$(dy)/dx =(1+2ycot(x - y) csc^2(x - y))/(cot^2(x-y)+2ycot(x - y) csc^2(x - y))$

How do you implicitly differentiate -1=x-ycot^2(x-y) -1=x-ycot^2(x-y) ?