Derivative of Composite Function

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Theorem

Let $f, g, h$ be continuous real functions such that:

$\forall x \in \R: h \left({x}\right) = f \circ g \left({x}\right) = f \left({g \left({x}\right)}\right)$


Then:

$h' \left({x}\right) = f' \left({g \left({x}\right)}\right) g' \left({x}\right)$

where $h'$ denotes the derivative of $h$.


Using the $D_x$ notation:

$D_x \left({f \left({g \left({x}\right)}\right)}\right) = D_{g \left({x}\right)} \left({f \left({g \left({x}\right)}\right)}\right) D_x \left({g \left({x}\right)}\right)$

This is often informally referred to as the chain rule (for differentiation).


Corollary

$\displaystyle \frac {\mathrm d y}{\mathrm d x} = \frac {\left({\dfrac {\mathrm d y}{\mathrm d u} }\right)}{\left({\dfrac {\mathrm d x}{\mathrm d u} }\right) }$

for $\dfrac {\mathrm d x}{\mathrm d u} \ne 0$.


Second Derivative

$D_x^2 w = D_u^2 w \left({D_x^1 u}\right)^2 + D_u^1 w D_x^2 u$


Third Derivative

$D_x^3 w = D_u^3 w \left({D_x^1 u}\right)^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$


Proof

Let $g \left({x}\right) = y$, and let:

\(\displaystyle g \left({x + \delta x}\right)\) \(=\) \(\displaystyle y + \delta y\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \delta y\) \(=\) \(\displaystyle g \left({x + \delta x}\right) - g \left({x}\right)\) $\quad$ $\quad$


Thus:

$\delta y \to 0$ as $\delta x \to 0$

and:

$\dfrac {\delta y} {\delta x} \to g' \left({x}\right) \qquad (1)$


There are two cases to consider:


Case 1

Suppose $g' \left({x}\right) \ne 0$ and that $\delta x$ is small but non-zero.

Then $\delta y \ne 0$ from $(1)$ above, and:

\(\displaystyle \lim_{\delta x \to 0} \frac {h \left({x + \delta x}\right) - h \left({x}\right)} {\delta x}\) \(=\) \(\displaystyle \lim_{\delta x \to 0} \frac {f \left({g \left({x + \delta x}\right)}\right) - f \left({g \left({x}\right)}\right)} {g \left({x + \delta x}\right) - g \left({x}\right)} \frac {g \left({x + \delta x}\right) - g \left({x}\right)} {\delta x}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\delta x \to 0} \frac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \frac {\delta y} {\delta x}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f' \left({y}\right) g' \left({x}\right)\) $\quad$ $\quad$

hence the result.

$\Box$


Case 2

Now suppose $g' \left({x}\right) = 0$ and that $\delta x$ is small but non-zero.

Again, there are two possibilities:


Case 2a

If $\delta y = 0$, then $\dfrac {h \left({x + \delta x}\right) - h \left({x}\right)} {\delta x} = 0$.

Hence the result.

$\Box$


Case 2b

If $\delta y \ne 0$, then $\dfrac {h \left({x + \delta x}\right) - h \left({x}\right)} {\delta x} = \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \dfrac {\delta y} {\delta x}$.


As $\delta y \to 0$:

$(1): \quad \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \to f' \left({y}\right)$
$(2): \quad \dfrac {\delta y} {\delta x} \to 0$


Thus:

$\displaystyle \lim_{\delta x \to 0} \frac {h \left({x + \delta x}\right) - h \left({x}\right)} {\delta x} \to 0 = f^{\prime} \left({y}\right) g^{\prime} \left({x}\right)$

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 $\left({\dfrac{\mathrm d y}{\mathrm d x}}\right)$ allows for a particularly elegant statement of this rule:

$\dfrac{\mathrm d y}{\mathrm d x} = \dfrac{\mathrm d y}{\mathrm d u} \cdot \dfrac{\mathrm d u}{\mathrm d x}$

where:

$\dfrac {\mathrm d y} {\mathrm d x}$ is the derivative of $y$ with respect to $x$
$\dfrac {\mathrm d y} {\mathrm d u}$ is the derivative of $y$ with respect to $u$
$\dfrac {\mathrm d u} {\mathrm d x}$ 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