Chain Rule for Real-Valued Functions

Theorem
Let $f: \R^n \to \R, \mathbf x \mapsto z$ be a differentiable real-valued function.

Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \R^n$.

Further, let every element $x_i: 1 \le i \le n$ represent an implicitly defined differentiable real function of $t$.

Then $z$ is itself differentiable $t$ and:

where $\dfrac {\partial z} {\partial x_i}$ is the partial derivative of $z$ $x_i$.

Proof
By hypothesis, $f$ is differentiable.

From Characterization of Differentiability:

Let $\Delta t \ne 0$ and divide both sides of the equation by $\Delta t$:

Recall that each $x_i$ was defined to be differentiable $t$, that is, that each $\dfrac {\d x_i} {\d t}$ exists.

Then $\Delta x_i \to 0$ as $\Delta t \to 0$.

Therefore:

Also see

 * Chain Rule for Derivatives
 * Definition:Total Derivative
 * Definition:Exact Differential Equation