# What is the slope of xy^3+xy=10 at the point (5,1)?

Jul 13, 2017

Slope $= - \frac{1}{10}$ at $\left(5 , 1\right)$

#### Explanation:

$x {y}^{3} + x y = 10$

Applying implicit differentiation:

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}}$ represents the slope of the tangent to the curve at any point $\left({x}_{1} , {y}_{1}\right)$ on the curve.

Slope of the tangent $\left(m\right)$ at $\left(5 , 1\right)$

$= \frac{- \left(1 + 1\right)}{3 \cdot 5 \cdot 1 + 5} = \frac{- 2}{15 + 5} = - \frac{2}{20}$

$= - \frac{1}{10}$

Mar 30, 2018

$\text{slope } = - \frac{1}{10}$

#### Explanation:

$\text{differentiate "color(blue)"implicitly with respect to x}$

$\text{differentiate "xy^3" and "xy" using the "color(blue)"product rule}$

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

$\Rightarrow 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{3} + x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{y}^{3} + y}{3 x {y}^{2} + x}$

$\text{at } \left(5 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1 + 1}{15 + 5} = - \frac{2}{20} = - \frac{1}{10} \leftarrow \textcolor{red}{\text{slope}}$