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

Jul 18, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{20}{21}$

#### 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:
$\frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)$

This one isn't too hard; you just have to apply the chain rule and power rule:
$\frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)$
$\to 2 \cdot 3 \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$= 6 y \frac{\mathrm{dy}}{\mathrm{dx}}$

Now, onto $4 x y$. We will need the power, chain, and product rules for this one:
$\frac{d}{\mathrm{dx}} \left(4 x y\right)$
$\to 4 \frac{d}{\mathrm{dx}} \left(x y\right)$
$= 4 \left(\left(x\right) ' \left(y\right) + \left(x\right) \left(y\right) '\right) \to$ Product rule: $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
$= 4 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$= 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Alright, finally ${x}^{2} y$ (more product, power, and chain rules):
$\frac{d}{\mathrm{dx}} \left({x}^{2} y\right)$
$= \left({x}^{2}\right) ' \left(y\right) + \left({x}^{2}\right) \left(y\right) '$
$= 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

Now that we have found all of our derivatives, we can express the problem as:
$\frac{d}{\mathrm{dx}} \left(3 {y}^{2} + 4 x y + {x}^{2} y\right) = \frac{d}{\mathrm{dx}} \left(C\right)$
$\to 6 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
(Remember the derivative of a constant is $0$).

Now we collect terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side and move everything else to the other:
$6 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\to 6 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(4 y + 2 x y\right)$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} \left(6 y + 4 x + {x}^{2}\right) = - \left(4 y + 2 x y\right)$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 y + 2 x y}{6 y + 4 x + {x}^{2}}$

All that's left to do is plug in $\left(2 , 5\right)$ to find our answer:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 y + 2 x y}{6 y + 4 x + {x}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 \left(5\right) + 2 \left(2\right) \left(5\right)}{6 \left(5\right) + 4 \left(2\right) + {\left(2\right)}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{20 + 20}{30 + 8 + 4}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{40}{42} = - \frac{20}{21}$