Question

What is the slope of the tangent line of `xy-y^2+x =C`, where C is an arbitrary constant, at `(-3,1)`?

Answer 1

`4/7`

Explanation:

`xy-y^2+x =C`

`(xy)' -(y^2)'+(x)' =0`

`(y + xy') -(2y y')+1 =0`

`y' (x -2y) =-(1 + y)`

`y' =(1 + y)/(2y-x)`

`y'_{(-1,3)} = (1 + 3)/(2(3) - (-1)) = 4/7`