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:
\(\ds \map g {x + \delta x}\) | \(=\) | \(\ds y + \delta y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \delta y\) | \(=\) | \(\ds \map g {x + \delta x} - \map g x\) |
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:
- $\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \, \map {g'} x$
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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 3$. Functions of Several Variables
- 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
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: differentiation (vi)