Derivative of Composite Function/Informal Proof

Theorem

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:

$\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \map {D_x} {\map g x}$

Some sources, introducing the Derivative of Composite Function at elementary level, provide the following non-rigorous argument:

If $z$ is a function of $y$ where $y$ itself is some function of $x$,

it is obvious that:

$\dfrac {\delta z} {\delta x} = \dfrac {\delta z} {\delta y} \cdot \dfrac {\delta y} {\delta x}$

since these quantities being finite can be cancelled as in Arithmetic.

If we now let the quantities concerned tend to zero, taking the limit, we get

$\dfrac {\d z} {\d x} = \dfrac {\d z} {\d y} \cdot \dfrac {\d y} {\d x}$

However, this informal argument is insufficiently rigorous for $\mathsf{Pr} \infty \mathsf{fWiki}$.

Hence, this must not be interpreted to mean that derivatives can be treated as fractions.

It simply is a convenient notation.

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Double Function