How do you implicitly differentiate csc(x^2+y^2)=e^(-xy) ?

1 Answer
Apr 8, 2016

y'=(-2ycsc(x^2+y^2)cot(x^2+y^2)+xe^(-xy))/(2x csc(x^2+y^2)cot(x^2+y^2)-ye^(-xy))

Explanation:

(-csc(x^2+y^2)cot(x^2+y^2))[2x+2yy']=e^(-xy)[-xy'-y]

-2x csc(x^2+y^2)cot(x^2+y^2) -2yy'csc(x^2+y^2)cot(x^2+y^2)=-xy'e^(-xy)-ye^(-xy)

-2yy'csc(x^2+y^2)cot(x^2+y^2)+xy'e^(-xy) = 2x csc(x^2+y^2)cot(x^2+y^2)-ye^(-xy)

y'(-2ycsc(x^2+y^2)cot(x^2+y^2)+xe^(-xy))=2x csc(x^2+y^2)cot(x^2+y^2)-ye^(-xy)

y'=(-2ycsc(x^2+y^2)cot(x^2+y^2)+xe^(-xy))/(2x csc(x^2+y^2)cot(x^2+y^2)-ye^(-xy))