How do you find the slope of the tangent to the curve y^3x+y^2x^2=6 at (2,1)?

Feb 20, 2015

Use implicit differentiation and the product rule

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

Do some rewriting

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

Factor and move terms without a $\frac{\mathrm{dy}}{\mathrm{dx}}$ factor to right side

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

now divide both sides by $3 x {y}^{2} + 2 {x}^{2} y$ and factor where you can

dy/dx=(-y^2(y+2x))/(yx(3y+2x)

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

Now evaluate at the given point $\left(2 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1 \left(1 + 2 \left(2\right)\right)}{2 \left(3 \left(1\right) + 2 \left(2\right)\right)} = \frac{- 1 \left(5\right)}{2 \left(7\right)} = \frac{- 5}{14}$