How do you implicitly differentiate #-y^2=e^(2x-4y)-2y/x #?

1 Answer

#y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)#

Explanation:

Start differentiating with respect to x both sides of the equation
#-y^2=e^(2x-4y) -2*y/x#

#d/dx(-y^2)=d/dx(e^(2x-4y)) -d/dx(2*y/x)#

#(-2y*y')=(e^(2x-4y)*d/dx(2x-4y)) -(2*d/dx(y/x))#

#-2y*y'=e^(2x-4y)(2-4y') -2*((xy'-y*1)/x^2)#

Simplify by dividing by 2

#-y*y'=e^(2x-4y)(1-2y') -((xy'-y*1)/x^2)#

#-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(xy')/x^2+(y)/x^2#

#-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(y')/x+y/x^2#

#2*e^(2x-4y)y' +(y')/x-y*y'=e^(2x-4y)+y/x^2#

Factor the y'

#(2*e^(2x-4y) +1/x-y)y'=e^(2x-4y)+y/x^2#

Divide both sides by #(2*e^(2x-4y) +1/x-y)#

#((2*e^(2x-4y) +1/x-y)y')/((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#

#cancel((2*e^(2x-4y) +1/x-y)y')/cancel((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#

#y'=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)#

Simplify by multiplying the right side by #x^2/x^2#

#y'=((e^(2x-4y)+y/x^2))/((2*e^(2x-4y) +1/x-y))*x^2/x^2#

#y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)#

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