Implicit Differentiation

2025. 1. 17. 18:02·Mathematics/Calculus

Theorem 1

Theorem 1. Suppose that $F(x, y)$ is differentiable and that the equation $F(x, y) = 0$ defines $y$ as a differentiable function of $x$. Then at any point where $\partial_y F \neq 0,$ $$\frac{dy}{dx} = - \frac{\partial_x F}{\partial_y F}.$$
Proof. Since $F(x, y) = 0$, the derivative $\frac{dF}{dx}$ must be zero. By the Chain Rule, we find $$0 = \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx} \\ \Longrightarrow \frac{dy}{dx} = - \frac{\partial_x F}{\partial_y F}. \blacksquare$$

Theorem 2

Theorem 2. Suppose that $F(x, y, z)$ is differentiable and that the equation $F(x, y, z) = 0$ defines $z$ as a differentiable function of $x$ and $y$. Then at any point where $\partial_z F \neq 0,$ $$\frac{\partial z}{\partial x} = - \frac{\partial_x F}{\partial_z F}, \\ \frac{\partial z}{\partial y} = - \frac{\partial_y F}{\partial_z F}.$$
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