# 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)?

Jan 29, 2016

I got $m = - \frac{3}{23}$

#### Explanation:

Given: $3 {y}^{2} + \frac{y}{x} + {x}^{2} / y = C$, and $\left(2 , 2\right)$ lies on the graph.

Find: the slope of the tangent line to the graph at the point $\left(2 , 2\right)$.

Solution:

$C = 3 {\left(2\right)}^{2} + \frac{2}{2} + {\left(2\right)}^{2} / 2 = 15$

So $3 {y}^{2} + \frac{y}{x} + {x}^{2} / y = 15$,

and $3 x {y}^{3} + {y}^{2} + {x}^{3} = 15 x y$.

Differentiating with respect to $x$ yields,

$3 {y}^{3} + 9 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {x}^{2} = 15 y + 15 x \frac{\mathrm{dy}}{\mathrm{dx}}$.

Therefore,

$\left(9 x {y}^{2} + 2 y - 15 x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 15 y - 3 {y}^{3} - 3 {x}^{2}$.

And, so,

$\left(9 \left(2\right) {\left(2\right)}^{2} + 2 \left(2\right) - 15 \left(2\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 15 \left(2\right) - 3 {\left(2\right)}^{3} - 3 {\left(2\right)}^{2}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{30 - 24 - 12}{72 + 4 - 30} = - \frac{6}{46} = - \frac{3}{23}$

Jan 29, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{23}$

#### Explanation:

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

$6 y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\frac{\mathrm{dy}}{\mathrm{dx}} \left(x\right) - y}{x} ^ 2 + \frac{2 x y - \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2}\right)}{y} ^ 2 = 0$

Plug in the point $\left(2 , 2\right)$ and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$12 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2}{4} + \frac{8 - 4 \frac{\mathrm{dy}}{\mathrm{dx}}}{4} = 0$

$48 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 + 8 - 4 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$46 \frac{\mathrm{dy}}{\mathrm{dx}} = - 6$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{23}$