# What is the slope of the tangent line of x^3y^2-(x+y)/(x-y)^2= C , where C is an arbitrary constant, at (1,4)?

Jan 20, 2018

Slope of tangent line is $- \frac{1303}{203}$

#### Explanation:

As we are seeking slope at $\left(1 , 4\right)$, this point lies on the curve given by ${x}^{3} {y}^{2} - \frac{x + y}{x - y} ^ 2 = C$.

Putting values from $\left(1 , 4\right)$, we get ${1}^{3} \cdot {4}^{2} - \frac{1 + 4}{1 - 4} ^ 2 = C$

or $C = 16 - \frac{5}{9} = \frac{139}{9}$ and function is ${x}^{3} {y}^{2} - \frac{x + y}{x - y} ^ 2 = \frac{139}{9}$

Now differentiating the imlicit function ${x}^{3} {y}^{2} - \frac{x + y}{x - y} ^ 2 = C$

$3 {x}^{2} {y}^{2} + 2 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1 + \frac{\mathrm{dy}}{\mathrm{dx}}}{x - y} ^ 2 - \frac{2 \left(x + y\right)}{x - y} ^ 3 \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

or $3 {x}^{2} {y}^{2} {\left(x - y\right)}^{3} + 2 {x}^{3} y {\left(x - y\right)}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - \left(x - y\right) \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 \left(x + y\right) \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

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

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

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

and at $\left(1 , 4\right)$, slope is $\frac{- 3 \cdot {1}^{2} \cdot {4}^{2} {\left(1 - 4\right)}^{3} + 3 + 4}{2 \cdot {1}^{3} \cdot 4 {\left(1 - 4\right)}^{3} + 1 + 3 \cdot 4}$

= $\frac{48 \cdot 27 + 7}{- 8 \cdot 27 + 13}$

= $- \frac{1303}{203}$

The tangent at $\left(1 , 4\right)$ looks liike:

graph{(x^3y^2-(x+y)/(x-y)^2-139/9)(203y-812+1303x-1303)=0 [-10.87, 9.13, -4.08, 5.92]}