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

##### 1 Answer
Nov 4, 2016

slope = $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{13}{11}$

#### Explanation:

To find the slope of the tangent line we have to find the derivative of the function first.

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

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

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

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

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

Now plug in 1 for x and -1 for y

dy/dx=(3(1--1)^2--1)/(3(1--1)^2+2(-1)+1

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{13}{11}$