How to differentiate compositions of functions.

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Concept |
Content |

Composition |
The composition for maps is just applying two maps in a row. |

Differentiability |
How to quantify the rate of change of a function. |

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**Theorem 1** (Chain rule). *Let \(I,J\) be intervals and \(f:J\to \mathbb{R}\) and \(g:I\to J\) be given. Assume that for \(x_0\in I\), \(g\) is differentiable in \(x_0\) and \(f\) is differentiable in \(g(x_0)\). Then the composition \(f\circ g\) (\((f\circ g)(x_0)=f(g(x_0))\)) is
differentiable in \(x_0\) with \[(f\circ g)'(x_0)=f'(g(x_0))\cdot
g'(x_0).\]*

*Proof:* By assumption there are functions \(\Delta_{g,x_0}(x):I\rightarrow \mathbb{R}\)
and \(\Delta_{f,g(x_0)}(y):J\rightarrow
\mathbb{R}\) which are continuous in \(x_0\) and \(g(x_0)\) respectively such that \[\begin{aligned}
g(x) &=& g(x_0)+(x-x_0)\Delta_{g,x_0}(x) \\
f(y) &=& f(g(x_0))+(y-g(x_0))\Delta_{f,g(x_0)}(y)
\end{aligned}\] for all \(x\in
I\) and \(y\in J\). Thus \[\begin{aligned}
(f\circ g)(x) &=& f(g(x)) =
f(g(x_0))+(g(x)-g(x_0))\Delta_{f,g(x_0)}(g(x)) \\
&=&
f(g(x_0))+(x-x_0)\Delta_{g,x_0}(x)\Delta_{f,g(x_0)}(g(x_0)+(x-x_0)\Delta_{g,x_0}(x))
\end{aligned}\] for all \(x\in
I\). Hence the function \[\begin{aligned}
\Delta_{f\circ g}(x) &:=&
\Delta_{g,x_0}(x)\Delta_{f,g(x_0)}(g(x_0)+(x-x_0)\Delta_{g,x_0}(x))
\end{aligned}\] fulfills \[\begin{aligned}
(f\circ g)(x) &=& (f\circ g)(x_0))+(x-x_0)\Delta_{f\circ g}(x)
\end{aligned}\] for all \(x\in
I\). As a composition of functions that are continuous in \(x_0\) it is also continuous in \(x_0\) with \(\Delta_{f\circ g}(x_0)=
\Delta_{g,x_0}(x_0)\Delta_{f,g(x_0)}(g(x_0))
=g'(x_0)f'(g(x_0))\). \(\Box\)

**Example 2**.

\(f:\mathbb{R}\to\mathbb{R}\)
with \(f(x)=\sin(x^2)\), then \(f'(x)=\cos(x^2)\cdot 2x\).

Let \(a\in\mathbb{R}\). Consider
\(f:\mathbb{R}\to\mathbb{R}\) with
\(f(x)=a^x=\exp(x\log(a))\), then \[f'(x)=\exp(x\log(a))\cdot\log(a)=a^x\cdot\log(a).\]

Let \(a\in\mathbb{R}\). Consider
\(f:\mathbb{R}^+\to\mathbb{R}\) with
\(f(x)=x^a=\exp(a\log(x))\), then \[f'(x)=\exp(a\log(x))\cdot\frac{a}x=x^a\cdot
\frac{a}x=a\cdot x^{a-1}.\]

Consider \(f:\mathbb{R}^+\to\mathbb{R}\) with \(f(x)=x^x=\exp(x\log(x))\), then \[f'(x)=\exp(x\log(x))\cdot\left(\frac{x}x+\log(x)\right)=x^x\cdot
(1+\log(x)).\]

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