## Test 2

Test 2, worth 15% and based on Chapter 3, will now take place Week 12, 30 April. There is a sample test [to give an indication of length and layout only] in the notes (marking scheme) and the test will be based on Chapter 3 only.

More Q. 1s (on the test) can be found on p.112; more Q. 2s on p. 117; more Q. 3s on p.125 and p.172, Q.1; more Q. 4s on p.136, and more Q. 5s on p. 143.

Chapter 3 Summary p. 144.

Please feel free to ask me questions via email or even better on this webpage.

## Homework

Once you are prepared for Test 2 you can start looking at Chapter 4:

- Revision of Integration, p.161.
- p.167, Q. 1-5
- p.182

## Week 11

We had some tutorial time from 18:00-19:00 for further differentiation (i.e. for Test 2, after Easter).

We completed our review of antidifferentiation before starting Chapter 4 proper.

We looked at Integration by Parts and centroids.

For those who could not make it here is some video and slides from what we did after the video died.

## Week 12

We will have some tutorial time from 18:00-19:00 for further differentiation.

We will have Test 2 from 19:00-20:05.

Then we will look at completing the square, centres of gravity, and work.

## Week 13

We will look the Winter 2018 paper at the back of your manual.

## CIT Mathematics Exam Papers

These are not always found in your programme selection — most of the time you will have to look here.

## Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

## 2 comments

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April 11, 2019 at 7:16 am

StudentHi J.P.,

Stuck on p. 102, Q 1 (f)

I have attached my attempt. I’ve been checking out khan academy and it said I needed the change of base formula but I got confused from there.

Thanks.

April 11, 2019 at 7:23 am

J.P. McCarthyThe confusion here, largely my fault perhaps, is that when you see in the calculus context that it is actually the natural log. So that, in this context:

.

This is the confusion.

So what you are differentiating is . It isn’t and so you have a chain rule where the ‘outside’ is and the ‘inside’ is .

The chain rule says that

,

that is “differentiate the outside — evaluate that at the inside — and the multiply by the derivative of the inside’.

The derivative of the outside, is in the tables and is . Evaluating this at the inside gives but we must multiply this by the derivative of the inside, the derivative of , which is .

All in all we have:

.

You probably need to move beyond this revision and tackle the new material — it might be the case that you don’t have to calculate more difficult derivatives such as these to do the new material.

Regards,

J.P.