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

Mar 3, 2016

The slope is $0$

#### Explanation:

$5 {x}^{3} {y}^{2} - y = C$ at $\left(0 , 0\right)$

$15 {x}^{2} {y}^{2} + 10 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} - 1 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

At $\left(0 , 0\right)$, we get

$0 + 0 - 1 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, so $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$