Question

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

Answer 1

`dy/dx=y/(x-2xy-x^2e^y)`

Explanation:

As such functions cannot be explicitly written in the form `y=f(x)`, we differentiate them implicitly w.r.t `x` or `y`, but when we differentiate w.r.t. `y`, we multiply by `dy/dx` too (using chain rule).

Hence, differential of `-y^2=e^y-y/x` is

`-2y dy/dx=e^y dy/dx-(x*dy/dx-y*1)/x^2` or

`-2y dy/dx=e^y dy/dx-1/x*dy/dx+y/x^2` or

`1/x*dy/dx-2y dy/dx-e^y dy/dx=y/x^2` or

`(1/x-2y-e^y)dy/dx=y/x^2` or

`dy/dx=y/(x^2(1/x-2y-e^y))=y/(x-2xy-x^2e^y)`