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

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

$3$ Functions

Derivative of Composite Function/3 Functions

Proof

This article contains statements that are justified by handwavery. In particular: Very neat derivation, but it's not really ProofWiki-standard rigour You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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.

$\blacksquare$

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.

Examples

Example: $\paren {3 x + 1}^2$

$\map {\dfrac \d {\d x} } {\paren {3 x + 1}^2} = 6 \paren {3 x + 1}$

Example: $\map \sin {x^2}$

$\map {\dfrac \d {\d x} } {\map \sin {x^2} } = 2 x \map \cos {x^2}$

Example: $\sqrt {1 + x}$

$\map {\dfrac \d {\d x} } {\sqrt {1 + x} } = \dfrac 1 {2 \sqrt {1 + x} }$

Example: $\sqrt {\sin x}$

$\map {\dfrac \d {\d x} } {\sqrt {\sin x} } = \dfrac {\cos x} {2 \sqrt {\sin x} }$

Also see

• Chain Rule for Real-Valued Functions

• Faà di Bruno's Formula

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Double Function
• 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 $21$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Entry: chain rule (for differentiation)
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Entry: chain rule (for differentiation)
• 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)