# Chain Rule for Real-Valued Functions

## Contents

## 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.

From Characterization of Differentiability:

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

$\blacksquare$

## Also see

## Sources

- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards:
*Calculus*(8th ed.): Appendix $A$: Theorem $13.6$