# How do you find the slope of a tangent line to the graph of the function x^3 + y^3 – 6xy = 0, at (4/3, 8/3)?

Apr 5, 2018

$\frac{4}{5}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + {y}^{3} - 6 x y\right) = \frac{d}{\mathrm{dx}} \left(0\right)$

$3 {x}^{2} + 3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 6 \left(1\right) y - 6 x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$, which gives the tangent slope at point $\left(x , y\right)$

$3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} + 6 y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 6 x\right) = - 3 {x}^{2} + 6 y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2} + 6 y}{3 {y}^{2} - 6 x}$

plug in $\left(\frac{4}{3} , \frac{8}{3}\right)$:

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(\frac{4}{3} , \frac{8}{3}\right) = \frac{- 3 {\left(\frac{4}{3}\right)}^{2} + 6 \left(\frac{8}{3}\right)}{3 {\left(\frac{8}{3}\right)}^{2} - 6 \left(\frac{4}{3}\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(\frac{4}{3} , \frac{8}{3}\right) = \frac{- \frac{16}{3} + 16}{\frac{64}{3} - 8}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(\frac{4}{3} , \frac{8}{3}\right) = \frac{\frac{32}{3}}{\frac{40}{3}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(\frac{4}{3} , \frac{8}{3}\right) = \frac{4}{5}$