1 Chain Rule
1.1 Chain Rule for Composition of Functions
In Calculus the chain rule provides a formula for the derivative of the composition of two differentiable real functions.
Consider differentiable real functions \(f(x)\) and \(g(x)\), where \(g\) is defined on an open interval containing \(a\) and \(f\) is defined on an open interval containing \(g(a)\). The derivative of the composite function \(h(x)=(f\circ g)(x)=f(g(x))\) is given by the chain rule \[ h'(x)=f'(g(x))\cdot g'(x). \] Written in an alternate but equivalent form, if we have \(y=f(u)\) and \(u=g(x)\) then the derivative of \(y\) is given by \[ \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}. \]
PROOF: Without loss of generality we assume \(f,g:\mathbb{R} \to \mathbb{R}\). We first define an auxiliary function \(\phi:\mathbb{R} \to \mathbb{R}\) as follows: \[ \phi(t)=\begin{cases}\frac{{f(t)-f(g(a))}}{t-g(a)} & \text{if }t \neq g(a) \\f'(g(a)) & \text{if }t=g(a).\end{cases} \] By construction, \(\phi\) is continuous at \(g(a)\). Moreover \(g\) is continuous at \(a\) because it is differentiable at \(a\). Hence \(\phi \circ g\) is continuous at \(a\). Note that for all \(x \neq a\) \[ \frac{{f(g(x))-f(g(a))}}{x-a}=\phi(g(x))\cdot \frac{{g(x)-g(a)}}{x-a}. \] Hence we have that \[ \begin{align}(f\circ g)'(a) & =\lim_{ x \to a } \frac{{f(g(x))-f(g(a))}}{x-a} \\& = \lim_{ x \to a } \phi(g(x))\cdot \lim_{ x \to a } \frac{{g(x)-g(a)}}{x-a} \\& = \phi(g(a))\cdot g'(a) \\& = f'(g(a))\cdot g'(a).\end{align} \] \(\square\)
1.2 Chain Rule for Higher Order Compositions
The chain rule can be easily applied to higher-order compositions by iterative application. For example, we consider the function \(y = e^{\sin(x^2)}\) which can be expressed as \(y=(f\circ g\circ h)(x)\) where
\[
\begin{align}
& f(x)=e^x,\quad g(x)=\sin(x)\quad \&\quad h(x)=x^2 \\
\implies & f'(x)=e^x,\quad g'(x)=\cos(x)\quad \& \quad h'(x)=2x.
\end{align}
\] From recursive application of the chain rule we have that \[
\begin{align}
(f\circ g\circ h)'(x) & =(f'\circ g\circ h)(x)\cdot (g\circ h)'(x) \\
& = (f'\circ g \circ h)(x)\cdot (g'\circ h)(x)\cdot h'(x).
\end{align}
\] Thus our final result is \[
(f\circ g\circ h)'(x)=e^{\sin(x^2)}\cdot \cos(x^2)\cdot 2x.
\] For \(n\)-length compositions of functions \(f_{1}, \dots, f_{n}\) we can summarize the above result as \[
(f_{1}\circ\dots\circ f_{n})=f'_{n}(x)\cdot\prod_{i=1}^{n-1}(f_{i}'\circ f_{i+1}\circ \dots \circ f_{n})(x).
\]
1.3 Quotient Rule
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