Editing Chain rule (section)

== Applications ==
[[File:Chain rule en.png|thumb|upright=1.6|The chain rule in case of composites of more than two functions]]
=== 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>

=== Quotient rule ===
{{See also|Quotient rule}}
The chain rule can be used to derive some well-known differentiation rules. For example, the quotient rule is a consequence of the chain rule and the [[product rule]]. To see this, write the function {{math|''f''(''x'')/''g''(''x'')}} as the product {{math|''f''(''x'') · 1/''g''(''x'')}}. First apply the product rule:
<math display="block">\begin{align}
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)
&= \frac{d}{dx}\left(f(x)\cdot\frac{1}{g(x)}\right) \\
&= f'(x)\cdot\frac{1}{g(x)} + f(x)\cdot\frac{d}{dx}\left(\frac{1}{g(x)}\right).
\end{align}</math>

To compute the derivative of {{math|1/''g''(''x'')}}, notice that it is the composite of {{mvar|g}} with the reciprocal function, that is, the function that sends {{mvar|x}} to {{math|1/''x''}}. The derivative of the reciprocal function is <math>-1/x^2\!</math>. By applying the chain rule, the last expression becomes:
<math display="block">f'(x)\cdot\frac{1}{g(x)} + f(x)\cdot\left(-\frac{1}{g(x)^2}\cdot g'(x)\right)
= \frac{f'(x) g(x) - f(x) g'(x)}{g(x)^2},</math>
which is the usual formula for the quotient rule.

=== Derivatives of inverse functions ===
{{Main|Inverse functions and differentiation}}
Suppose that {{math|1=''y'' = ''g''(''x'')}} has an [[inverse function]]. Call its inverse function {{mvar|f}} so that we have {{math|1=''x'' = ''f''(''y'')}}. There is a formula for the derivative of {{mvar|f}} in terms of the derivative of {{mvar|g}}. To see this, note that {{mvar|f}} and {{mvar|g}} satisfy the formula
<math display="block">f(g(x)) = x.</math>

And because the functions <math>f(g(x))</math> and {{mvar|x}} are equal, their derivatives must be equal. The derivative of {{mvar|x}} is the constant function with value 1, and the derivative of <math>f(g(x))</math> is determined by the chain rule. Therefore, we have that:
<math display="block">f'(g(x)) g'(x) = 1.</math>

To express {{mvar|f'}} as a function of an independent variable {{mvar|y}}, we substitute <math>f(y)</math> for {{mvar|x}} wherever it appears. Then we can solve for {{mvar|f'}}.
<math display="block">\begin{align}
f'(g(f(y))) g'(f(y)) &= 1 \\
f'(y) g'(f(y)) &= 1 \\
f'(y) = \frac{1}{g'(f(y))}.
\end{align}</math>

For example, consider the function {{math|1=''g''(''x'') = ''e''<sup>''x''</sup>}}. It has an inverse {{math|1=''f''(''y'') = ln ''y''}}. Because {{math|1=''g''′(''x'') = ''e''<sup>''x''</sup>}}, the above formula says that
<math display="block">\frac{d}{dy}\ln y = \frac{1}{e^{\ln y}} = \frac{1}{y}.</math>

This formula is true whenever {{mvar|g}} is differentiable and its inverse {{mvar|f}} is also differentiable. This formula can fail when one of these conditions is not true. For example, consider {{math|1=''g''(''x'') = ''x''<sup>3</sup>}}. Its inverse is {{math|1=''f''(''y'') = ''y''<sup>1/3</sup>}}, which is not differentiable at zero. If we attempt to use the above formula to compute the derivative of {{mvar|f}} at zero, then we must evaluate {{math|1=1/''g''′(''f''(0))}}. Since {{math|1=''f''(0) = 0}} and {{math|1=''g''′(0) = 0}}, we must evaluate 1/0, which is undefined. Therefore, the formula fails in this case. This is not surprising because {{mvar|f}} is not differentiable at zero.

=== Back propagation ===
The chain rule forms the basis of the [[back propagation]] algorithm, which is used in [[gradient descent]] of [[neural network (machine learning)|neural networks]] in [[deep learning]] ([[artificial intelligence]]).<ref>{{citation|title=Deep learning|first1=Ian|last1=Goodfellow|authorlink1=Ian Goodfellow|first2=Yoshua|last2=Bengio|authorlink2=Yoshua Bengio|first3=Aaron|last3=Courville|year=2016|publisher=MIT}}, pp=197–217.</ref>