# Derivative of Composite Function

## Contents

## 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

- $D_x^2 w = D_u^2 w \left({D_x^1 u}\right)^2 + D_u^1 w D_x^2 u$

### Third Derivative

- $D_x^3 w = D_u^3 w \left({D_x^1 u}\right)^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$

## 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.

## Also see

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 13$: General Rules of Differentiation: $13.11$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 10.13$ - 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $21$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**chain rule**