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

Question

1520 views around the world

You can reuse this answer
Creative Commons License

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))#