Theorem 1
Theorem 1. If $w=f(x, y)$ is differentiable and if $x = x(t), y=y(t)$ are differentiable functions of $t$, then the composition $w=f(x(t), y(t))$ is a differentiable function of $t$ and $$\frac{dw}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}.$$
Proof. Let $\Delta x, \Delta y$ and $\Delta w$ be the increments that result from changing $t$ from $t_0$ to $t_0 + \Delta t$. Since $f$ is differentiable, $$\Delta w = \frac{\partial f}{\partial x}(x(t_0), y(t_0)) \Delta x + \frac{\partial f}{\partial y}(x(t_0), y(t_0)) \Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y, $$ where $\varepsilon_1, \varepsilon_2 \to 0$ as $\Delta x, \Delta y \to 0$. Let's divide this equation through by $\Delta t$ and let $\Delta t$ approach zero. Then the division gives $$\frac{\Delta w}{\Delta t} = \frac{\partial f}{\partial x}(x(t_0), y(t_0)) \frac{\Delta x}{\Delta t} + \frac{\partial f}{\partial y}(x(t_0), y(t_0)) \frac{\Delta y}{\Delta t} + \varepsilon_1 \frac{\Delta x}{\Delta t} + \varepsilon_2 \frac{\Delta y}{\Delta t}.$$ Letting $\Delta t$ approach zero gives $$\frac{dw}{dt}(t_0) = \lim_{\Delta t \to 0} \frac{\Delta w}{\Delta t} \\ = \frac{\partial f}{\partial x}(x(t_0), y(t_0)) \frac{d x}{d t}(t_0) + \frac{\partial f}{\partial y}(x(t_0), y(t_0)) \frac{d y}{d t}(t_0) + 0 \cdot \frac{d x}{d t}(t_0) + 0 \cdot \frac{d y}{d t}(t_0). \blacksquare$$
위 정리의 합성 함수에서 $x, y$는 $f$의 'intermediate variable'이고, $t$야말로 independent variable이라고 할 수 있다. 이러한 관계를 아래와 같은 diagram으로 그릴 수 있는데, 이를 dependency diagram이라고 부른다. 기억할 때는 diagram의 각 line을 따라 곱해주고 모두 더해주면 된다.
Theorem 2
Theorem 2. If $w=f(x, y, z)$ is differentiable and $x, y$, and $z$ are differentiable functions of $t$, then $w$ is a differentiable function of $t$ and $$\frac{dw}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}.$$
three-variable의 경우에도 Theorem 1과 동일하다.
Theorem 3
Theorem 3. Suppose that $w = f(x, y, z), x = g(r, s), y = h(r, s)$ and $z = k(r, s)$. If all four functions are differentiable, then $w$ has partial derivatives with respect to $r$ and $s$, given by the formulas $$\frac{\partial w}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial r} \\ \frac{\partial w}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial s}.$$
Theorem 4
Theorem 4. Suppose that $w = f(x_1, x_2, ..., x_n)$ is a differentiable function of the $n$ variables $x_1, x_2, ..., x_n$ and each one is a differentiable function of the $m$ variables $t_1, t_2, ..., t_m$. Then $w$ is a differentiable function of $t_1, t_2, ..., t_m$ and $$\frac{\partial w}{\partial t_i} = \frac{\partial f}{\partial x_1} \frac{\partial x_1}{\partial t_i} + \frac{\partial f}{\partial x_2} \frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial f}{\partial x_n} \frac{\partial x_n}{\partial t_i}$$ for each $i = 1, 2, ..., m$.
위와 같이 Chain Rule을 일반화 하여 작성할 수 있고, $$\left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, ..., \frac{\partial f}{\partial x_n} \right) \cdot \left( \frac{\partial x_1}{\partial t_i}, \frac{\partial x_2}{\partial t_i}, ..., \frac{\partial x_n}{\partial t_i} \right)$$와 같은 mnemonic으로, 즉 두 벡터의 내적으로 쉽게 기억할 수 있다. 이때 왼쪽 벡터는 사실상 $\nabla f$이다.