Derivative of Composite Function

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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


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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


Sources

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