# How do you use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy= 2 at the point ( 1 , 1 )?

Jun 26, 2015

Use implicit differentiation to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, eveluate $\frac{d}{\mathrm{dx}}$ at $\left(\frac{1}{1}\right)$ to get the slope of the tangent line. Find the equation of the line.

#### Explanation:

$x {y}^{3} + x y = 2$

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

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

Both terms on the left are products, so we'll use the product rule. I use the order: $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$

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

Simplify to get:

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

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

At $\left(1 , 1\right)$, the slope of the tangent line is $= \frac{1}{2}$ and the equation of the line is:

$y = - \frac{1}{2} x + \frac{3}{2}$.