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

1 Answer

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

Explanation:

Start differentiating both sides of the equation with respect to #x#

#2=y-(e^(2y))/x#

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

#0=y'-[x*d/dx(e^(2y))-e^(2y)*d/dx(x)]/x^2#

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

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

Transpose the terms with #y'# to the left side of the equation

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

Factor out #y'#

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

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

Divide both sides by #((2e^(2y)-x)/x)#

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

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

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