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:

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

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:

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


Proof

$f$ is by hypothesis differentiable.

From Characterization of Differentiability:

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

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

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

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

$\blacksquare$


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