# What is the slope of the tangent line of 1/x^3-y^2-y = C , where C is an arbitrary constant, at (-5,0)?

Nov 22, 2015

The slope is $\frac{- 3}{625}$

#### Explanation:

The constant $C$
If we know that $\left(- 5 , 0\right)$ is a point on the graph, then $C = - \frac{1}{125}$.
So $C$ is determined by the values of $x$ and $y$. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

$\frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 3 - {y}^{2} - y\right) = \frac{d}{\mathrm{dx}} \left(C\right)$

$- \frac{3}{x} ^ 4 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- \frac{3}{x} ^ 4 = \left(2 y + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

The slope at $\left(- 5 , 0\right)$, is therefore

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{625}$