Question

What is the slope of the tangent line of `3y^2+4xy+x^2y =C`, where C is an arbitrary constant, at `(2,5)`?

Answer 1

`dy/dx=-20/21`

Explanation:

You will need to know the basics of implicit differentiation for this problem.

We know the slope of the tangent line at a point is the derivative; so the first step will be to take the derivative. Let's do it piece by piece, starting with:
`d/dx(3y^2)`

This one isn't too hard; you just have to apply the chain rule and power rule:
`d/dx(3y^2)`
`->2*3*y*dy/dx`
`=6ydy/dx`

Now, onto `4xy`. We will need the power, chain, and product rules for this one:
`d/dx(4xy)`
`->4d/dx(xy)`
`=4((x)'(y)+(x)(y)')->` Product rule: `d/dx(uv)=u'v+uv'`
`=4(y+xdy/dx)`
`=4y+4xdy/dx`

Alright, finally `x^2y` (more product, power, and chain rules):
`d/dx(x^2y)`
`=(x^2)'(y)+(x^2)(y)'`
`=2xy+x^2dy/dx`

Now that we have found all of our derivatives, we can express the problem as:
`d/dx(3y^2+4xy+x^2y)=d/dx(C)`
`->6ydy/dx+4y+4xdy/dx+2xy+x^2dy/dx=0`
(Remember the derivative of a constant is `0`).

Now we collect terms with `dy/dx` on one side and move everything else to the other:
`6ydy/dx+4y+4xdy/dx+2xy+x^2dy/dx=0`
`->6ydy/dx+4xdy/dx+x^2dy/dx=-(4y+2xy)`
`->dy/dx(6y+4x+x^2)=-(4y+2xy)`
`->dy/dx=-(4y+2xy)/(6y+4x+x^2)`

All that's left to do is plug in `(2,5)` to find our answer:
`dy/dx=-(4y+2xy)/(6y+4x+x^2)`
`dy/dx=-(4(5)+2(2)(5))/(6(5)+4(2)+(2)^2)`
`dy/dx=-(20+20)/(30+8+4)`
`dy/dx=-(40)/(42)=-20/21`