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

1 Answer
Bdub
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))#

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