# 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 with respect to $t$ and:

 $\displaystyle \frac {\d z} {\d t}$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \frac {\partial z }{\partial x_i} \frac {\d x_i} {\d t}$

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

### Corollary

Let $\Psi$ represent a differentiable function of $x$ and $y$.

Let $y$ represent a differentiable function of $x$.

Then:

 $\displaystyle \frac {\d \Psi} {\d x}$ $=$ $\displaystyle \frac {\partial \Psi} {\partial x} + \frac {\partial \Psi} {\partial y} \frac {\d y} {\d x}$

## Proof

By hypothesis, $f$ is differentiable.

 $\displaystyle \Delta z$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \epsilon_i \Delta x_i$ $\forall i: 1 \le i \le n: \epsilon_i \to 0$ as $\Delta x_i \to 0$

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

 $\displaystyle \frac {\Delta z} {\Delta t}$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\Delta x_i} {\Delta t} + \sum_{i \mathop = 1}^n \epsilon_i \frac {\Delta x_i} {\Delta t}$ $\forall i: 1 \le i \le n: \epsilon_i \to 0$ as $\Delta x_i \to 0$

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

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

Therefore:

 $\displaystyle \frac {\d z} {\d t}$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\d x_i} {\d t} + \sum_{i \mathop = 1}^n 0 \frac {\d x_i} {\d t}$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\d x_i} {\d t}$ as $\Delta t \to 0$

$\blacksquare$