Question

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

Answer 1

I got `m = -3/23`

Explanation:

Given: `3y^2+y/x+x^2/y =C`, and `(2,2)` lies on the graph.

Find: the slope of the tangent line to the graph at the point `(2,2)`.

Solution:

`C = 3(2)^2+2/2+(2)^2/2 = 15`

So `3y^2+y/x+x^2/y =15`,

and `3xy^3+y^2+x^3 = 15xy`.

Differentiating with respect to `x` yields,

`3y^3+9xy^2 dy/dx + 2y dy/dx + 3x^2 = 15y+15x dy/dx`.

Therefore,

`(9xy^2 + 2y - 15x) dy/dx = 15y-3y^3- 3x^2`.

And, so,

`(9(2)(2)^2 + 2(2) - 15(2)) dy/dx = 15(2)-3(2)^3- 3(2)^2`.

`dy/dx = (30-24-12)/(72+4-30) = -6/46 = -3/23`

Answer 2

`dy/dx=-3/23`

Explanation:

This can also be done without evaluating the constant and using quotient rule:

`6ydy/dx+(dy/dx(x)-y)/x^2+(2xy-dy/dx(x^2))/y^2=0`

Plug in the point `(2,2)` and solve for `dy/dx`.

`12dy/dx+(2dy/dx-2)/4+(8-4dy/dx)/4=0`

`48dy/dx+2dy/dx-2+8-4dy/dx=0`

`46dy/dx=-6`

`dy/dx=-3/23`