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The sum, product and quotient rules show us how to differentiate a great many different functions from the reals to the reals. However some functions, such as $f(x)=\sin 2x$ are a composition of functions, and these rules don’t tell us what the derivative of $\sin 2x$ is. There is, however, a theorem called the chain rule that tells us how to differentiate these functions. Here we present the proof. In class we won’t prove this assertion but we will make one attempt to explain why it takes the form it does. In general only practise can make you proficient in the use of the chain rule. See http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf or any other textbook (such as a LC text book) with exercises.

Proposition 4.1.8 (Chain Rule)

Let $f, g:\mathbb{R}\rightarrow\mathbb{R}$ be functions, and let $F$ denote the composition $F=g\circ f$ (that is $F(x)=g(f(x))$ for each $x\in\mathbb{R}$). If $a\in\mathbb{R}$ such that $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $F$ is differentiable at $a$ with

$F'(a)=g'(f(a))f'(a)$

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Let $f,g:\mathbb{R}\rightarrow \mathbb{R}$. Their composition is the function $g\circ f:\mathbb{R}\rightarrow \mathbb{R}$ is defined by

$(g\circ f)(x)=g(f(x))$

Let $f$ and $g$ be functions $\mathbb{R}\rightarrow \mathbb{R}$ with $f$ continuous at some point $a\in\mathbb{R}$, and $g$ continuous at the point $f(a)$. Then $g\circ f$ is continuous at $a$.

Proof: For each $\varepsilon>0$, we must find a $\delta>0$ such that

$|x-a|<\delta\Rightarrow |(g\circ f)(x)-(g\circ f)(a)|<\varepsilon$

Let $\varepsilon>0$, since $g$ is continuous at $f(a)$, $\exists\,\delta_g>0$:

$|t-f(a)|<\delta_g\Rightarrow |g(t)-g(f(a))|<\varepsilon$

But also $f$ is continuous at $a$, so (we can get $f(x)$ $\delta_g$-close to $f(a)$), $\exists\,\delta>0$ such that

$|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g$

So therefore,

$|x-a|<\delta\Rightarrow |f(x)-f(a)|<\delta_g\Rightarrow |g(f(x))-g(f(a))|<\varepsilon$

$\Box$

Here we present the proof of assertions 3. and 4. of the following proposition. The proofs of 1. and 2. will be presented in class and here they are assumed. The proofs presented here will not be presented in class.

Proposition 3.1.4 (Calculus of Limits)
Suppose that $f$ and $g$ are two functions $\mathbb{R}\rightarrow \mathbb{R}$, and that for some $a\in\mathbb{R}$ we have

$\lim_{x\rightarrow a}f(x)=p$ , and  $\lim_{x\rightarrow a}g(x)=q$.

for some $p,q\in\mathbb{R}$. Then

1. $\lim_{x\rightarrow a} (f(x)+g(x))=p+q$.
2. If $k\in\mathbb{R}$, $\lim_{x\rightarrow a} kf(x)=kp$.
3. $\lim_{x\rightarrow a} (f(x)g(x))=pq$.
4. If $q\neq0$, $\lim_{x\rightarrow a} (f(x)/g(x))=p/q$.
5. If $n\in\mathbb{N}$, and $p>0$ then $\lim_{x\rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{p}$.
6. Read the rest of this entry »