Part 3/6:
Here, ( \frac{dy}{dx} ) represents the derivative of y with respect to x. Rearranging this equation leads us to find ( \frac{dy}{dx} ):
[
2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}
]
This derivative is significant as it describes the slope of the implicit function at any point without ever needing to isolate y.
Finding Derivatives at Specific Points
To find the derivative at a specific point, substitute the x-coordinate into the original implicit equation to find the corresponding y-value. For example, if we wish to find the derivative at ( x = 3 ) in the equation ( x^2 + y^2 = 25 ), we substitute:
[
3^2 + y^2 = 25 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4
]