Variables Bound Together
Sometimes you cannot solve for \(y\). For example: \(x^2 + y^2 = 25\) (a circle). To find the slope \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) and using the Chain Rule.
Worked Examples
Example 1: The Circle
Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = 25\).
- Differentiate \(x^2 \to 2x\).
- Differentiate \(y^2\) using Chain Rule: \(2y \cdot \frac{dy}{dx}\).
- Differentiate \(25 \to 0\).
- Equation: \(2x + 2y \frac{dy}{dx} = 0\).
- Solve for \(\frac{dy}{dx}\): \(2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}\).
- Result: \(\frac{dy}{dx} = -\frac{x}{y}\).
Example 2: Mixed Terms
Find \(\frac{dy}{dx}\) for \(x^2 y + y^2 x = 6\).
- Term 1 (\(x^2 y\)): Use Product Rule \(\to (2x)(y) + (x^2)(\frac{dy}{dx})\).
- Term 2 (\(y^2 x\)): Use Product Rule \(\to (2y \frac{dy}{dx})(x) + (y^2)(1)\).
- Combine: \(2xy + x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y^2 = 0\).
- Group \(\frac{dy}{dx}\): \(\frac{dy}{dx}(x^2 + 2xy) = -2xy - y^2\).
- Result: \(\frac{dy}{dx} = -\frac{2xy + y^2}{x^2 + 2xy}\).
The Bridge to Quantum Mechanics
In classical mechanics, position and momentum are independent. But in Quantum Mechanics, they are "coupled" by the Uncertainty Principle. When we look at Phase Space trajectories (where position and momentum are coordinates), we often find equations where \(x\) and \(p\) cannot be separated. Implicit differentiation allows us to find the "flow" of a quantum system through this abstract space even when we can't solve for one variable explicitly.