I am emailing a link of this to everyone on the class list every Wednesday morning. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

## [EDIT] Definition Questions

If you really want to get good at your definition questions check out these (tough) exercises..

## Next Week

We will have an additional tutorial instead of a lecture on Monday 24.

## This Week

On Monday, we covered from (but not including) Definition 3.1.1 to (and including) Corollary 3.1.6. On Wednesday we went through Questions 1 & 2 from 2010/11’s Test 1 A and we did Q. 3 from the sample test.

In the tutorial we answered exercises Q. 1 (vii), 3 (b), 4(ii) & 8(iii) from Exercise Sheet 2.

## Problems

You need to do exercises – all of the following you should be able to attempt. Do as many as you can/ want in the following order of most beneficial:

### Wills’ Exercise Sheets

Q. 1, 2, 3(i) & (iii), 4, 8(a) & (b) from Exercise Sheet 3.

### More Exercise Sheets

Nothing from from Problems.

### Past Exam Papers

Nothing from Summer 2010.

Nothing from Autumn 2010.

Nothing from Summer 2009.

Nothing from Autumn 2009.

Q. 4(a) from Summer 2008.

Nothing from Autumn 2008.

Nothing from Summer 2007.

Q. 4(b) from Autumn 2007.

Nothing from Summer 2006.

Nothing from Autumn 2006.

Nothing from Summer 2005.

Nothing from Autumn 2005.

Nothing from Summer 2004.

Nothing from Autumn 2004.

Nothing from Summer 2003.

Nothing from Autumn 2003.

Q. 6(a) from Summer 2002.

Q. 4(b) from Summer 2001.

Q. 4(b) from Summer 2000.

### From the Class

1. Prove Proposition 3.1.4 (ii).
2. Prove Corollary 3.1.6.

## Supplementary Notes

We never proved the following in class.

### Proposition 3.14

Let $f,\,g:\mathbb{R}\rightarrow\mathbb{R}$ be functions that are differentiable at some $a\in\mathbb{R}$Then

(iii) $fg$ is differentiable at $a$ with $(fg)'(a)=f'(a)g(a)+g'(a)f(a)$ [Product Rule]

(iv) if $g(a)\neq 0$then $f/g$ is differentiable at $a$ with

$\left(\frac{f}{g}\right)'(a)=\frac{g(a)f'(a)-f(a)g'(a)}{[g(a)]^2}$  [Quotient Rule]

Proof:

(iii)

$(fg)'(a)=\lim_{h\rightarrow 0}\frac{fg(a+h)-fg(a)}{h}$

$=\lim_{h\rightarrow 0}\frac{f(a+h)g(a+h)\overbrace{-f(a+h)g(a)+f(a+h)g(a)}^{=0}-f(a)g(a)}{h}$

$=\lim_{h\rightarrow 0}\frac{f(a+h)[g(a+h)-g(a)]+g(a)[f(a+h)-f(a)]}{h}$

$=\lim_{h\rightarrow 0}\left(\underbrace{f(a+h)}_{\rightarrow f(a)}\times \underbrace{\frac{g(a+h)-g(a)}{h}}_{\rightarrow g'(a)}+\underbrace{g(a)}_{\rightarrow g(a)}\times\underbrace{\frac{f(a+h)-f(a)}{h}}_{\rightarrow g'(a)}\right)$

(iv) Let $q=f/g$:

$q(a+h)-q(a)=\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}$

$=\frac{f(a+h)g(a)-f(a)g(a+h)}{g(a+h)g(a)}$

$=\frac{f(a+h)g(a)\overbrace{-f(a)g(a)+f(a)g(a)}^{=0}-f(a)g(a+h)}{g(a+h)g(a)}$

$=\frac{g(a)[f(a+h)-f(a)]-f(a)[g(a+h)-g(a)]}{g(a+h)g(a)}$
$\Rightarrow \frac{q(a+h)-q(a)}{h}=\frac{g(a)\left[\frac{f(a+h)-f(a)}{h}\right]-f(a)\left[\frac{g(a+h)-g(a)}{h}\right]}{g(a+h)g(a)}$

Letting $h\rightarrow 0$ on both sides:

$q'(a)=\left(\frac{f}{g}\right)'(a)=\frac{g(a)f'(a)-f(a)g'(a)}{[g(a)]^2}\,\,\,\bullet$