# 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)?

##### 1 Answer
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 $\left(1 , 2\right)$:

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

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