Chain Rule for Real-Valued Functions

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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:

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

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:

\(\displaystyle \frac {\mathrm d \Psi} {\mathrm d x}\) \(=\) \(\displaystyle \frac {\partial \Psi} {\partial x} + \frac {\partial \Psi} {\partial y} \frac {\mathrm d y} {\mathrm d x}\)


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 $\dfrac {\mathrm d x_i} {\mathrm d t}$ exists.

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

Therefore:

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

$\blacksquare$


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