Derivative of Composite Function
Theorem
Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
- $\forall x \in \R: \map h x = \map {f \circ g} x = \map f {\map g x}$
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:
- $\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \map {D_x} {\map g x}$
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
Let $f, g, h$ be continuous real functions such that:
\(\ds y\) | \(=\) | \(\ds \map f u\) | ||||||||||||
\(\ds u\) | \(=\) | \(\ds \map g v\) | ||||||||||||
\(\ds h\) | \(=\) | \(\ds \map h x\) |
Then:
- $\dfrac {\d y} {\d x} = \dfrac {\d y} {\d u} \cdot \dfrac {\d u} {\d v} \cdot \dfrac {\d v} {\d x}$
Jacobians
Let $U$ be an open subset of $\R^n$.
Let $\mathbf f = \paren {f_1, f_2, \ldots, f_m}^\intercal: U \to \R^m$ be a vector valued function, differentiable at $\mathbf x = \paren {x_1, x_2, \ldots, x_n}^\intercal \in U$.
Let $\mathbf g = \paren {g_1, g_2, \ldots, g_k}^\intercal: U \to \R^k$ be a vector valued function, differentiable at $\map {\mathbf f} {\mathbf x} = \paren {\map {f_1} {\mathbf x}, \map {f_2} {\mathbf x}, \ldots, \map {f_m} {\mathbf x} }^\intercal \in U$.
The Jacobian matrix of a composite function $\mathbf g \circ \mathbf f$ is obtained by multiplying the Jacobian matrices of $\mathbf f$ and $\mathbf g$:
- $\map {\mathbf J_{\mathbf g \circ \mathbf f} } {\mathbf x} = \map {\mathbf J_{\mathbf g} } {\map {\mathbf f} {\mathbf x} } \map {\mathbf J_{\mathbf f} } {\mathbf x}$
Also presented as
The Derivative of Composite Function is also often seen presented using Leibniz's notation for derivatives:
- $\dfrac {\d y} {\d x} = \dfrac {\d y} {\d u} \cdot \dfrac {\d u} {\d x}$
or:
- $\dfrac \d {\d x} \map u v = \dfrac {\d u} {\d v} \cdot \dfrac {\d v} {\d x}$
where $\dfrac {\d y} {\d x}$ denotes the derivative of $y$ with respect to $x$.
Some sources go so far as to mix their notation and present something like this:
- $y' = \dfrac {\d f} {\d g} \map {g'} x$
Also to be mentioned is:
- $D_x^1 w = D_u^1 w \, D_x^1 u$
where ${D_x}^k u$ denotes the $k$th derivative of $u$ with respect to $x$.
Also known as
The rule to calculate the Derivative of Composite Function is often informally referred to as:
Some sources refer to it as just the Chain Rule, but the latter term has a number of applications.
Proof 1
Let $f, g, x_0$ satisfy the conditions of the theorem.
Define $g^* : I \to \R$ as:
- $\map {g^*} x = \begin{cases} \dfrac {\map g x - \map g {x_0}} {x - x_0} & : x \ne x_0 \\ \map {g'} {x_0} & : x = x_0 \end{cases}$
Then, for every $x \in I$:
- $\paren 1: \quad \map g x - \map g {x_0} = \paren {x - x_0} \map {g^*} x$
for when $x = x_0$, both sides are equal to $0$.
Furthermore, by the definition of derivative, it follows that $g^*$ is continuous at $x_0$.
Define $y_0 := \map g {x_0}$
In the same way, define $f^* : J \to \R$ as:
- $\map {f^*} y = \begin{cases} \dfrac {\map f y - \map f {y_0}} {y - y_0} & : y \ne y_0 \\ \map {f'} {y_0} & : y = y_0 \end{cases}$
Likewise, for every $y \in J$:
- $\paren 2: \quad \map f y - \map f {y_0} = \paren {y - y_0} \map {f^*} y$
and $f^*$ is continuous at $y_0$.
Now, for every $x \in I$, we have:
\(\ds \map f {\map g x} - \map f {\map g {x_0} }\) | \(=\) | \(\ds \map f {\map g x} - \map f {y_0}\) | Definition of $y_0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map g x - y_0} \map {f^*} {\map g x}\) | By $\paren 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map g x - \map g {x_0} } \map {f^*} {\map g x}\) | Definition of $y_0$ | |||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {x - x_0} \map {g^*} x \map {f^*} {\map g x}\) | By $\paren 1$ |
Therefore, we have:
\(\ds \lim_{x \mathop \to x_0} \frac {\map h x - \map h {x_0} } {x - x_0}\) | \(=\) | \(\ds \lim_{x \mathop \to x_0} \frac {\map f {\map g x} - \map f {\map g {x_0} } } {x - x_0}\) | Definition of $h$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to x_0} \map {g^*} x \map {f^*} {\map g x}\) | By $\paren 3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\lim_{x \mathop \to x_0} \map {g^*} x} \paren {\lim_{x \mathop \to x_0} \map {f^*} {\map g x} }\) | Product Rule for Limits of Real Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\lim_{x \mathop \to x_0} \map {g^*} x} \map {f^*} {\lim_{x \mathop \to x_0} \map g x}\) | Limit of Composite Function: Hypothesis 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {g^*} {x_0} \map {f^*} {\map g {x_0} }\) | Differentiable Function is Continuous, Definition of Continuous Real Function at Point | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {g'} {x_0} \map {f'} {\map g {x_0} }\) | Definitions of $g^*$ and $f^*$ |
Therefore, by definition of derivative:
- $\map {h'} {x_0} = \map {f'} {\map g {x_0}} \map {g'} {x_0}$
$\blacksquare$
Proof 2
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.
$\blacksquare$
Informal Proof
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.
- it is obvious that:
- 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.
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
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 3$. Functions of Several Variables
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): composite function (function of a function)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): composite function (function of a function)
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): chain rule
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): differentiation (vi)
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- 1957: Tom M. Apostol: Mathematical Analysis ... (previous): Chapter $5$: Differentiation of Functions of One Real Variable: $\text {5-4}$ The chain rule
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 10.13$