Derivative of Composite Function

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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:

the Chain Rule for Derivatives
the Chain Rule for Differentiation.

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.
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


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