# How do you use implicit differentiation to find the slope of the curve given xy^2+x^2y=2 at (1,-2)?

##### 2 Answers
Dec 31, 2017

Treat $y$ as a function of $x$, differentiate, solve for $y '$, and plug in the point to get a slope of 0.

#### Explanation:

Assuming the equation defines $y$ "implicitly" as a function of $x$, when we differentiate both sides of $x {y}^{2} + {x}^{2} y = 2$ with respect to $x$, using the Product Rule and Chain Rule, we get:

${y}^{2} + 2 x y y ' + 2 x y + {x}^{2} y ' = 0$.

This can be rearranged to $y ' \left(2 x y + {x}^{2}\right) = - {y}^{2} - 2 x y$, or

$y ' = \frac{- {y}^{2} - 2 x y}{2 x y + {x}^{2}}$

Plugging in the point $\left(x , y\right) = \left(1 , - 2\right)$ (which is on the curve since $1 \cdot {\left(- 2\right)}^{2} + {1}^{2} \cdot \left(- 2\right) = 4 - 2 = 2$) gives a slope of

$y ' = \frac{- {\left(- 2\right)}^{2} - 2 \cdot 1 \cdot \left(- 2\right)}{2 \cdot 1 \cdot \left(- 2\right) + {1}^{2}} = \frac{0}{- 3} = 0$.

If it helps, conceptually-speaking, you could write $f \left(x\right)$ in place of $y$ and write the original equation as $x {\left(f \left(x\right)\right)}^{2} + {x}^{2} f \left(x\right) = 2$ before differentiating.

Since the original equation is quadratic in $y$, you could also use the quadratic formula to find an "explicit" formula for $y$ in terms of $x$ for this example before differentiating.

Dec 31, 2017

See below

#### Explanation:

This particular equation will use the product and chain rule. When we differentiate implicitly, we use the idea of the chain rule when we differentiate $y$. This is based on the idea that $y$ is still a function of $x$ even though it is not given explicitly. So for example when we differentiate $y$ in the following, we find:

$\frac{\mathrm{dy}}{\mathrm{dy}} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

The $\frac{\mathrm{dy}}{\mathrm{dx}}$ is the derivative of $x$. This follows the same idea when using the chain rule to differentiate explicitly.We multiply by the inner function. Here we don't no what the inner function is, so we leave it in the form $\frac{\mathrm{dy}}{\mathrm{dx}}$. This is generally easier to grasp once you start using it.

So from example:

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

Apply product and chain rule:

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

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

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

Plugging in our coordinates:

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

GRAPH: 