Question

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

Answer 1

`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.