Editing Chain rule (section)

=== Composites of more than two functions ===
The chain rule can be applied to composites of more than two functions. To take the derivative of a composite of more than two functions, notice that the composite of {{mvar|f}}, {{mvar|g}}, and ''{{mvar|h}}'' (in that order) is the composite of {{mvar|f}} with {{math|''g'' ∘ ''h''}}. The chain rule states that to compute the derivative of {{math|''f'' ∘ ''g'' ∘ ''h''}}, it is sufficient to compute the derivative of ''{{mvar|f}}'' and the derivative of {{math|''g'' ∘ ''h''}}. The derivative of {{mvar|f}} can be calculated directly, and the derivative of {{math|''g'' ∘ ''h''}} can be calculated by applying the chain rule again.{{citation needed|date=November 2023}}

For concreteness, consider the function
<math display="block">y = e^{\sin (x^2)}.</math>
This can be decomposed as the composite of three functions:
<math display="block">\begin{align}
y &= f(u) = e^u, \\
u &= g(v) = \sin v, \\
v &= h(x) = x^2.
\end{align}</math>
So that <math> y = f(g(h(x))) </math>.

Their derivatives are:
<math display="block">\begin{align}
\frac{dy}{du} &= f'(u) = e^u, \\
\frac{du}{dv} &= g'(v) = \cos v, \\
\frac{dv}{dx} &= h'(x) = 2x.
\end{align}</math>

The chain rule states that the derivative of their composite at the point {{math|1=''x'' = ''a''}} is:
<math display="block">\begin{align}
(f \circ g \circ h)'(a) & = f'((g \circ h)(a)) \cdot (g \circ h)'(a) \\
& = f'((g \circ h)(a)) \cdot g'(h(a)) \cdot h'(a) \\
& = (f' \circ g \circ h)(a) \cdot (g' \circ h)(a) \cdot h'(a).
\end{align}</math>

In [[Leibniz's notation]], this is:
<math display="block">\frac{dy}{dx} = \left.\frac{dy}{du}\right|_{u=g(h(a))}\cdot\left.\frac{du}{dv}\right|_{v=h(a)}\cdot\left.\frac{dv}{dx}\right|_{x=a},</math>
or for short,
<math display="block">\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}.</math>
The derivative function is therefore:
<math display="block">\frac{dy}{dx} = e^{\sin(x^2)}\cdot\cos(x^2)\cdot 2x.</math>

Another way of computing this derivative is to view the composite function {{math|''f'' ∘ ''g'' ∘ ''h''}} as the composite of {{math|''f'' ∘ ''g''}} and ''h''. Applying the chain rule in this manner would yield:
<math display="block">\begin{align}
(f \circ g \circ h)'(a) &= (f \circ g)'(h(a)) \cdot h'(a) \\
&= f'(g(h(a))) \cdot g'(h(a)) \cdot h'(a).
\end{align}</math>

This is the same as what was computed above. This should be expected because {{math|1=(''f'' ∘ ''g'') ∘ ''h'' = ''f'' ∘ (''g'' ∘ ''h'')}}.

Sometimes, it is necessary to differentiate an arbitrarily long composition of the form <math>f_1 \circ f_2 \circ \cdots \circ f_{n-1} \circ f_n\!</math>. In this case, define
<math display="block">f_{a\,.\,.\,b} = f_{a} \circ f_{a+1} \circ \cdots \circ f_{b-1} \circ f_{b}</math>
where <math>f_{a\,.\,.\,a} = f_a</math> and <math>f_{a\,.\,.\,b}(x) = x</math> when <math>b < a</math>. Then the chain rule takes the form
<math display="block">\begin{align}
Df_{1\,.\,.\,n}
&= (Df_1 \circ f_{2\,.\,.\,n}) (Df_2 \circ f_{3\,.\,.\,n}) \cdots (Df_{n-1} \circ f_{n\,.\,.\,n}) Df_n \\
&= \prod_{k=1}^n \left[Df_k \circ f_{(k+1)\,.\,.\,n}\right]
\end{align}</math>
or, in the Lagrange notation,
<math display="block">\begin{align}
f_{1\,.\,.\,n}'(x)
&= f_1' \left( f_{2\,.\,.\,n}(x) \right) \; f_2' \left( f_{3\,.\,.\,n}(x) \right) \cdots f_{n-1}' \left(f_{n\,.\,.\,n}(x)\right) \; f_n'(x) \\[1ex]
&= \prod_{k=1}^{n} f_k' \left(f_{(k+1\,.\,.\,n)}(x) \right)
\end{align}</math>