Question

How do you find the slope of the tangent to the curve `y^3x+y^2x^2=6` at `(2,1)`?

Answer 1

Use implicit differentiation and the product rule

`3y^2dy/dxx+y^3+2ydy/dxx^2+y^2 2x=0`

Do some rewriting

`3xy^2dy/dx+2x^2ydy/dx+y^3+2xy^2=0`

Factor and move terms without a `dy/dx` factor to right side

`dy/dx(3xy^2+2x^2y)=-y^3-2xy^2`

now divide both sides by `3xy^2+2x^2y` and factor where you can

`dy/dx=(-y^2(y+2x))/(yx(3y+2x)`

`dy/dx=(-y(y+2x))/(x(3y+3x))`

Now evaluate at the given point `(2,1)`

`dy/dx=(-1(1+2(2)))/(2(3(1)+2(2)))=(-1(5))/(2(7))=(-5)/14`