Answer
At $(10,3), \frac{dy}{dx} = \frac{1}{6}$
At $(10,-3), \frac{dy}{dx} = -\frac{1}{6}$
Work Step by Step
First method:
We solve for y and then take the derivative:
$y^2-x+1 = 0$
$y = \pm \sqrt{x-1}$
$\frac{dy}{dx} = \pm \frac{1}{2} (x-1)^{-\frac{1}{2}}$
Using this formula, we get:
At $(10,3), \frac{dy}{dx} = \frac{1}{6}$ (because the slope is upward in the first quadrant, we take the positive $\frac{dy}{dx}$ value).
At $(10,-3), \frac{dy}{dx} = -\frac{1}{6}$ (because the slope is downward in the fourth quadrant, we take the negative $\frac{dy}{dx}$ value).
Second Method:
We use implicit differentiation.
$y^2 -x+1= 0$
$2y\frac{dy}{dx}-1 = 0$
$\frac{dy}{dx} = \frac{1}{2y}$
Using this formula, we get:
At $(10,3), \frac{dy}{dx} = \frac{1}{6}$
At $(10,-3), \frac{dy}{dx} = -\frac{1}{6}$
