# How do you use Implicit differentiation find x^2+ 2xy- y^2 + x=2 and to find an equation of the tangent line to the curve, at the point (1,2)?

Mar 30, 2015

I'll include more steps than you might want to write when you get more experience, but here we go:

${x}^{2} + 2 x y - {y}^{2} + x = 2$ Differentiate both sides with respect to $x$.

$\frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x y - {y}^{2} + x\right) = \frac{d}{\mathrm{dx}} \left(2\right)$ (Write this every time)

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

Remember that y is the name of some function(s) of $x$ that I haven't found expressions for.
So $2 x y$ is really $2 x \cdot \left(\text{some f of x}\right)$ We'll need the product and chain rules.(I use the product rule written as $\left(F S\right) ' = F ' S + F S '$.)
Similarly ${y}^{2}$ is really ${\left(\text{some f of x}\right)}^{2}$ so we'll need the power and chain rules. (Implicit differentiation is using the chain rule.)

$2 x \frac{\mathrm{dx}}{\mathrm{dx}} + \left[\frac{d}{\mathrm{dx}} \left(+ 2 x\right) \cdot y + \left(+ 2 x\right) \frac{d}{\mathrm{dx}} \left(y\right)\right] - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$2 x + 2 y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

We could solve for $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \left(2 x + 2 y + 1\right)}{2 x - 2 y}$, but if we don't have to, it's usually easier to substitute numbers now:

At $\left(1 , 2\right)$, we get:

$2 \left(1\right) + 2 \left(2\right) + 2 \left(1\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \left(2\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$ so

$2 + 4 + 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 4 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$7 - 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ and finally:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{2}$

The tangent line contains the point $\left(1 , 2\right)$ and has slope $m = \frac{7}{2}$ so its equation is:

$y = \frac{7}{2} x - \frac{3}{2}$

With experience, your solution will look more like:
${x}^{2} + 2 x y - {y}^{2} + x = 2$

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

$2 x + 2 y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

At $\left(1 , 2\right)$:

$2 + 4 + 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 4 \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{7}{2}$