# Question #6bfa5

Oct 23, 2017

(a) $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} + 5 y}{5 x + 3 {y}^{2}}$
(b) $y = - \frac{37}{27} x - \frac{19}{9}$

#### Explanation:

(a) Differentiate both sides of the equation.
Take notice that $\frac{d}{\mathrm{dx}} {y}^{3}$ = $\frac{d}{\mathrm{dy}} {y}^{3} \cdot \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$ (Chain Rule)
and $\frac{d}{\mathrm{dx}} x y$ = $y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ (Product Rule).

$\frac{d}{\mathrm{dx}} \left({x}^{3} + {y}^{3} + 5 x y\right) = \frac{d}{\mathrm{dx}} 65$
$3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 5 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\left(5 x + 3 {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} - 5 y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} + 5 y}{5 x + 3 {y}^{2}}$.

(b) Substitute $x = 3 , y = 2$ to $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} + 5 y}{5 x + 3 {y}^{2}}$.
The slope of the tangent line is $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 \cdot {3}^{2} + 5 \cdot 2}{5 \cdot 3 + 3 \cdot {2}^{2}} = - \frac{37}{27}$.

Therefore, the tangent line is:
$y - 2 = - \frac{37}{27} \left(x - 3\right)$
$y = - \frac{37}{27} x - \frac{19}{9}$.