Derivative of Composite Function

Theorem

Let $f, g, h$ be continuous real functions such that:

$\forall x \in \R: \map h x = \map {f \circ g} x = \map f {\map g x}$

Then:

$\map {h'} x = \map {f'} {\map g x} \map {g'} x$

where $h'$ denotes the derivative of $h$.

Using the $D_x$ notation:

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

This is often informally referred to as the chain rule (for differentiation).

Corollary

$\dfrac {\d y} {\d x} = \dfrac {\paren {\dfrac {\d y} {\d u} } } {\paren {\dfrac {\d x} {\d u} } }$

for $\dfrac {\d x} {\d u} \ne 0$.

Second Derivative

$D_x^2 w = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u$

Third Derivative

$D_x^3 w = D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$

Proof

Let $\map g x = y$, and let:

 $\displaystyle \map g {x + \delta x}$ $=$ $\displaystyle y + \delta y$ $\displaystyle \leadsto \ \$ $\displaystyle \delta y$ $=$ $\displaystyle \map g {x + \delta x} - \map g x$

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:

 $\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x}$ $=$ $\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map f {\map g {x + \delta x} } - \map f {\map g x} } {\map g {x + \delta x} - \map g x} \frac {\map g {x + \delta x} - \map g x} {\delta x}$ $\displaystyle$ $=$ $\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map f {y + \delta y} - \map f y} {\delta y} \frac {\delta y} {\delta x}$ $\displaystyle$ $=$ $\displaystyle \map {f'} y \, \map {g'} x$

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:

$\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \, \map {g'} x$

Again, hence the result.

$\Box$

All cases have been covered, so by Proof by Cases, the result is complete.

$\blacksquare$

Notation

Leibniz's notation for derivatives $\dfrac {\d y} {\d x}$ allows for a particularly elegant statement of this rule:

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

where:

$\dfrac {\d y} {\d x}$ is the derivative of $y$ with respect to $x$
$\dfrac {\d y} {\d u}$ is the derivative of $y$ with respect to $u$
$\dfrac {\d u} {\d x}$ is the derivative of $u$ with respect to $x$

However, this must not be interpreted to mean that derivatives can be treated as fractions. It simply is a convenient notation.