Derivative of Composite Function/Proof 1
Theorem
Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
- $\forall x \in \R: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for each $x_0 \in I$ such that:
- $g$ is differentiable at $x_0$
- $f$ is differentiable at $\map g {x_0}$
it holds that $h$ is differentiable at $x_0$ and:
- $\map {h'} {x_0} = \map {f'} {\map g {x_0}} \map {g'} {x_0}$
where $h'$ denotes the derivative of $h$.
Using the $D_x$ notation:
- $\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \map {D_x} {\map g x}$
Proof
Let $f, g, x_0$ satisfy the conditions of the theorem.
Define $g^* : I \to \R$ as:
- $\map {g^*} x = \begin{cases} \dfrac {\map g x - \map g {x_0}} {x - x_0} & : x \ne x_0 \\ \map {g'} {x_0} & : x = x_0 \end{cases}$
Then, for every $x \in I$:
- $\paren 1: \quad \map g x - \map g {x_0} = \paren {x - x_0} \map {g^*} x$
for when $x = x_0$, both sides are equal to $0$.
Furthermore, by the definition of derivative, it follows that $g^*$ is continuous at $x_0$.
Define $y_0 := \map g {x_0}$
In the same way, define $f^* : J \to \R$ as:
- $\map {f^*} y = \begin{cases} \dfrac {\map f y - \map f {y_0}} {y - y_0} & : y \ne y_0 \\ \map {f'} {y_0} & : y = y_0 \end{cases}$
Likewise, for every $y \in J$:
- $\paren 2: \quad \map f y - \map f {y_0} = \paren {y - y_0} \map {f^*} y$
and $f^*$ is continuous at $y_0$.
Now, for every $x \in I$, we have:
| \(\ds \map f {\map g x} - \map f {\map g {x_0} }\) | \(=\) | \(\ds \map f {\map g x} - \map f {y_0}\) | Definition of $y_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\map g x - y_0} \map {f^*} {\map g x}\) | By $\paren 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\map g x - \map g {x_0} } \map {f^*} {\map g x}\) | Definition of $y_0$ | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {x - x_0} \map {g^*} x \map {f^*} {\map g x}\) | By $\paren 1$ |
Therefore, we have:
| \(\ds \lim_{x \mathop \to x_0} \frac {\map h x - \map h {x_0} } {x - x_0}\) | \(=\) | \(\ds \lim_{x \mathop \to x_0} \frac {\map f {\map g x} - \map f {\map g {x_0} } } {x - x_0}\) | Definition of $h$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to x_0} \map {g^*} x \map {f^*} {\map g x}\) | By $\paren 3$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\lim_{x \mathop \to x_0} \map {g^*} x} \paren {\lim_{x \mathop \to x_0} \map {f^*} {\map g x} }\) | Product Rule for Limits of Real Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\lim_{x \mathop \to x_0} \map {g^*} x} \map {f^*} {\lim_{x \mathop \to x_0} \map g x}\) | Limit of Composite Function: Hypothesis 1 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {g^*} {x_0} \map {f^*} {\map g {x_0} }\) | Differentiable Function is Continuous, Definition of Continuous Real Function at Point | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {g'} {x_0} \map {f'} {\map g {x_0} }\) | Definitions of $g^*$ and $f^*$ |
Therefore, by definition of derivative:
- $\map {h'} {x_0} = \map {f'} {\map g {x_0}} \map {g'} {x_0}$
$\blacksquare$
Also presented as
The Derivative of Composite Function is also often seen presented using Leibniz's notation for derivatives:
- $\dfrac {\d y} {\d x} = \dfrac {\d y} {\d u} \cdot \dfrac {\d u} {\d x}$
or:
- $\dfrac \d {\d x} \map u v = \dfrac {\d u} {\d v} \cdot \dfrac {\d v} {\d x}$
where $\dfrac {\d y} {\d x}$ denotes the derivative of $y$ with respect to $x$.
Some sources go so far as to mix their notation and present something like this:
- $y' = \dfrac {\d f} {\d g} \map {g'} x$
Also to be mentioned is:
- $D_x^1 w = D_u^1 w \, D_x^1 u$
where ${D_x}^k u$ denotes the $k$th derivative of $u$ with respect to $x$.