# How do you use implicit differentiation to find the slope of the curve given xy^5+x^5y=1 at (-1,-1)?

Jul 26, 2016

Since $\left(- 1 , - 1\right)$ is not a point on the given curve, this question has no solution as it stands. However the slope of $x {y}^{5} + {x}^{5} y = 2$ at $\left(- 1 , - 1\right)$ is -1.

#### Explanation:

Differentiate both sides of the equation $x {y}^{5} + {x}^{5} y = 2$ with respect to $x$ leads to

$\frac{\mathrm{dx}}{\mathrm{dx}} {y}^{5} + x \frac{{\mathrm{dy}}^{5}}{\mathrm{dx}} + \frac{{\mathrm{dx}}^{5}}{\mathrm{dx}} y + {x}^{5} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or

${y}^{5} + 5 x {y}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {x}^{4} y + {x}^{5} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

So that at $\left(- 1 , - 1\right)$ we have

${\left(- 1\right)}^{5} + 5 \left(- 1\right) {\left(- 1\right)}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {\left(- 1\right)}^{4} \left(- 1\right) + {\left(- 1\right)}^{5} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

or

$- 6 \frac{\mathrm{dy}}{\mathrm{dx}} - 6 = 0$

so that the slope is -1.