# How do you find the slope of the tangent line to the curve xy^3 + xy = 4 at the point (2,1)? Thank you! I really, really, appreciate your help:)?

Sep 26, 2015

$s l o p e = - \frac{1}{4}$

#### Explanation:

$x {y}^{3} + x y = 4$
${y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + y + x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
slope at (2,1)
$1 + 6 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 + 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$8 \frac{\mathrm{dy}}{\mathrm{dx}} = - 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{4} \implies s l o p e = - \frac{1}{4}$