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

1 Answer

#y'=(-2x^2*csc (x^2+y^2)*cot (x^2+y^2)+e^(-x))/[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]#

Explanation:

#csc (x^2+y^2)=e^(-x)-y#

differentiate both sides of the equation with respect to x

#d/dx(csc (x^2+y^2))=d/dx(e^(-x)-y)#

#-csc (x^2+y^2)*cot (x^2+y^2)*d/dx(x^2+y^2)=d/dx(e^(-x))-d/dx(y)#

#-csc (x^2+y^2)*cot (x^2+y^2)(2x^2+2y*y')=(e^(-x))*(-1)-y'#

#-csc (x^2+y^2)*cot (x^2+y^2)(2x^2+2y*y')=-e^(-x)-y'#

Multiplication then transposition

#(2x^2(-csc (x^2+y^2)*cot (x^2+y^2))+e^(-x)=#
#2y*y'*(csc (x^2+y^2)*cot (x^2+y^2))-y'#

factor out the y'

#-2x^2csc (x^2+y^2)*cot (x^2+y^2)+e^(-x)=#
#[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]y'#

Divide both sides by #[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]#

#y'=(-2x^2*csc (x^2+y^2)*cot (x^2+y^2)+e^(-x))/[2y*csc (x^2+y^2)*cot (x^2+y^2)-1]#

God bless....I hope the explanation is useful