# How do you use implicit differentiation to find the slope of the curve given xy=-8 at (4,-2)?

Aug 1, 2016

The slope of tgt. at pt.$\left(4 , - 2\right) = \frac{1}{2}$.

#### Explanation:

Let us recall that $\frac{\mathrm{dy}}{\mathrm{dx}}$ gives the slope of tgt. at the pt. $\left(x , y\right)$.

Now, $x y = - 8$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(x y\right) = \frac{d}{\mathrm{dx}} \left(- 8\right) = 0$.

$\Rightarrow x \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{d}{\mathrm{dx}} \left(x\right) = 0$.

$\Rightarrow x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

$\therefore$ the slope of tgt. at pt. $\left(x , y\right) = - \frac{y}{x}$.

$\therefore$ the slope of tgt. at pt.$\left(4 , - 2\right) = - \frac{- 2}{4} = \frac{1}{2}$.