Working left to right, we have:

Derivative of `2y`:

`d/(dx)2y=2(dy)/(dx)`

Derivative of `5` is `0`.

Derivative of x2 is `2x`.

Derivative of y3:

`d/(dx)y^3=3y^2(dy)/(dx)`

Putting it together, implicit differentiation gives us:

`2(dy)/(dx)-2x-3y^2(dy)/(dx)=0`

Collecting like terms gives:

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

So

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

Now, when `x = 2` and `y = -1`,

`(dy)/(dx)=(2(2))/(2-3(-1)^2)`

`=4/-1`

`=-4`

So the slope of the tangent at `(2,-1)` is `-4`.

Let's see what we have done. We graph the curve

`2y+5-x^2-y^3=0`

and graph the tangent to the curve at `(2, -1)`. We see that indeed the slope is `-4`.

It works!

implicit graph