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

##### 1 Answer

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(1 , 4\right)} = \frac{35}{2}$

#### Explanation:

Since, the point $\left(1 , 4\right)$ lies on the curve: ${x}^{3} - \setminus \frac{x + y}{x - y} = C$ hence it will satisfy the equation of curve as follows

${1}^{3} - \setminus \frac{1 + 4}{1 - 4} = C$

$C = \frac{8}{3}$

hence, substituting above value, the equation of given curve is

${x}^{3} - \setminus \frac{x + y}{x - y} = \frac{8}{3}$

$y = \setminus \frac{3 {x}^{4} - 11 x}{3 {x}^{3} - 5}$

differentiating above equation w.r.t. $x$ using division rule, we get the slope

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\setminus \frac{3 {x}^{4} - 11 x}{3 {x}^{3} - 5}\right)$

$= \setminus \frac{\left(3 {x}^{3} - 5\right) \frac{d}{\mathrm{dx}} \left(3 {x}^{4} - 11 x\right) - \left(3 {x}^{4} - 11 x\right) \frac{d}{\mathrm{dx}} \left(3 {x}^{3} - 5\right)}{{\left(3 {x}^{3} - 5\right)}^{2}}$

$= \setminus \frac{\left(3 {x}^{3} - 5\right) \left(12 {x}^{3} - 11\right) - \left(3 {x}^{4} - 11 x\right) \left(9 {x}^{2}\right)}{{\left(3 {x}^{3} - 5\right)}^{2}}$

Hence, substituting $x = 1$ in above equation, the slope $\frac{\mathrm{dy}}{\mathrm{dx}}$ of tangent at $\left(1 , 4\right)$ is

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(1 , 4\right)} = \setminus \frac{\left(3 {\left(1\right)}^{3} - 5\right) \left(12 {\left(1\right)}^{3} - 11\right) - \left(3 {\left(1\right)}^{4} - 11 \setminus \cdot 1\right) \left(9 {\left(1\right)}^{2}\right)}{{\left(3 {\left(1\right)}^{3} - 5\right)}^{2}}$

$= \frac{35}{2}$