Derivative of Composite Function/Proof 2
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:
- $\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
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 $\map g x = y$, and let:
| \(\ds \map g {x + \delta x}\) | \(=\) | \(\ds y + \delta y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \delta y\) | \(=\) | \(\ds \map g {x + \delta x} - \map g x\) |
Thus:
- $\delta y \to 0$ as $\delta x \to 0$
and:
- $(1): \quad \dfrac {\delta y} {\delta x} \to \map {g'} x$
There are two cases to consider:
Case 1
Suppose $\map {g'} x \ne 0$ and that $\delta x$ is small but non-zero.
Then $\delta y \ne 0$ from $(1)$ above, and:
| \(\ds \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x}\) | \(=\) | \(\ds \lim_{\delta x \mathop \to 0} \frac {\map f {\map g {x + \delta x} } - \map f {\map g x} } {\map g {x + \delta x} - \map g x} \frac {\map g {x + \delta x} - \map g x} {\delta x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{\delta x \mathop \to 0} \frac {\map f {y + \delta y} - \map f y} {\delta y} \frac {\delta y} {\delta x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {f'} y \map {g'} x\) |
hence the result.
$\Box$
Case 2
Now suppose $\map {g'} x = 0$ and that $\delta x$ is small but non-zero.
Again, there are two possibilities:
Case 2a
If $\delta y = 0$, then $\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = 0$.
Hence the result.
$\Box$
Case 2b
If $\delta y \ne 0$, then:
- $\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta x}$
As $\delta y \to 0$:
- $(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$
- $(2): \quad \dfrac {\delta y} {\delta x} \to 0$
Thus:
- $\ds \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \map {g'} x$
Again, hence the result.
$\Box$
All cases have been covered, so by Proof by Cases, the result is complete.
$\blacksquare$
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:
- $D_x^1 w = D_u^1 w \, D_x^1 u$
or:
- ${D_x}^1 w = {D_u}^1 w \, {D_x}^1 u$
where ${D_x}^k u$ denotes the $k$th derivative of $u$ with respect to $x$.