Differentiating from first principles

This question comes up frequently (not every year) in the Leaving Cert. When it comes up, you usually have to differentiate a quadratic, though you should practice differentiating linear functions (e.g. f(x) = 4x -1) also. It is usually worth 20 marks.

The way I approach it is to set up the derivative function as if you were finding the slope using the slope formula m = (y2-y1)/(x2-x1).
Today's example which we did in class was

3x²-4x+2 (I think)

x1 = x
y1 = 3x²-4x+2
x2 = x + h
y2 = 3(x+h)²-4(x+h)+2

If you plug these into the slope formula and multiply out the y2 and watch your signs with y1, you'll get

m =
3x²+3h² +6xh -4x-4h+2 - 3x²+4x-2
------------------------------------------------
x + h - x

= 3h² +6xh -4h
---------------
h

= 3h +6x -4

Now apply the Limit as h tends towards zero (you'll lose marks if you don't do this part properly)

Lim [f(x+h) - f(x)]/h = 6x -4
h->0

Also, write down somewhere either f(x) = 3x²-4x+2 and f'(x) = 6x -4

or
y = 3x²-4x+2

dy
-- = 6x -4
dx

It is worth spending time practicing and understanding this process. You only have to do it if you are specifically asked to differentiate "from first principles", but it is central to figuring out other topics in functions and differentiation, and it gives you an application of coordinate geometry and practice at some tricky algebra.

I have a worksheet that I didn't have time to distribute today that gives a worked example and practice of differentiating from first principles.

We'll look at the rules for differentiating (product, quotient and chain rules) tomorrow.
Test on Friday.

Finishing Functions, Graphs and Differentiation.

We need to finish functions, graphs and our calculus.
In tomorrow's class we will look at differentiating from first principles and at how to apply the different rules (product, quotient and chain rules).
Everyone should do the 2007 Q6,7,8 and 1999 Q7,8 at least.

We'll have a class test on Thursday.

Functions and Graphs

We are working on functions and graphs at the moment.

In today's class we looked at how you can use a function to solve a related function.
Using graphs of functions to find the roots of a polynomial is a key skill.
When you graph a function, you find the output values for a selection of discreet input values.
When you join the dots with a smooth curve you are extrapolating the output values of the input values in between. That is why we graph functions.
Wherever your line crosses the x-axis, the output value is 0 - don't forget equation of the x-axis is y=0. So you can use these points to work out the roots of a cubic equation.
We looked at the 1999 q8 question on functions. This included working out a related function. Make sure you do this question and are able to follow it.
When you split the related cubic function into the original cubic and a related linear function, you can work out the roots of the related function by reading the points of intersection from your graph.

Other points on functions which you need to know are mentioned in the 1 page overview.

Algebra

We are spending a few days revising algebra. This is a really important topic as it comes up all over the paper 1 and 2 as well as in 2 algebra-only questions (Paper 1 Q2 and Q3).

We started by looking at the 2007 questions.
Q2 involved solving inequalities. Key points:
- know how to graph on the numberline or work out the solution set = { , , } if x is an element of N, Z and R
- know how to handle cases like -x <>3 ]

It also involved manipulating indices. It is too difficult to format the rules here... look them up in your book and make sure you know them. With equations, get the same number raised to different powers on each side of the =.
Eg 2^(x+1) = 4^x
rewrite 4^x as (2^2)^x or 2^2x
Then equate the powers:
x+1 = 2x
x = 1

We will look at the rest of Q2 and Q3 tomorrow. Here is a slightly revised version of the algebra 1-page overview.

More trigonometry

The unit circle, compound angles etc took longer than planned on Friday. As I said in class, these topics come up infrequently, so if you are aiming to pass this question, make sure that you are able to use sin=opp/hyp etc, sine rule, cosine rule and area of triangle.

I forgot to post the 1-page overview of trigonometry last time.



Also, I will mention this in tomorrow's class, but there is one thing you should be aware of when using the Sine rule:
If you use your calculator to work out the inverse sine of 0.9897, your calculator will tell you it is 81.7 degrees ... which it is. But, it is also the sine of 97.3 degrees. You know from the discussion of unit circle that there are two possible angles that have the same sine, cosine and tan. Our calculators always return the angle that is between 0 and 90 degrees only, not the obtuse alternative.
Look at 2003 P2 Q5 for an example where this can cause problems.
If you solve (c) (ii) by putting
Sin60/7 = SinA/8
you will get the wrong answer.
The problem here is that the shorter side of the triangle (3) is not used in this equation and so the triangle where we have a known angle (60) and two other sides (7 and 8) is not a well defined triangle. In effect, it is ASS (angle, side, side) rather than SAS (side, angle, side)! (remember the 4 conditions for congruence of triangles that you learned: SSS,SAS,ASA,RHS)
You have to use the acute angle to work it out.
Sin60/7 = SinB/3 ... which gives you 22 degrees for that angle and leaves 98 degrees for the obtuse angle (60 + 22 + 98 = 180).

How do you know which angle to go for? A triangle can have at most 1 obtuse angle (> 90 degrees) and that will be opposite the longest side.
If you have a choice, solve the acute angle, opposite the shortest side, first.

Trigonometry

There are two parts to the trigonometry course.

The first part relates to finding missing sides and angles or the area of a triangle:

• application of the trigonometric ratios for sin, cos and tan (sin=opp/hyp etc.) for right-angled triangles
• application of the sine rule for cases where you have one complete and one incomplete "angle and opposite side" pair
• application of cosine rule where you have 2 sides and the angle between them, or just 3 sides of a triangle
• application of area of triangle rule 1/2 ab SinC

This is relatively straightforward and covers about 90% of the questions that come up on the exam.

The second part relates to compound angles, angles > 90 degrees, compass directions, unit circle, handling special angles (30, 45 and 60 degrees) using surds. This is the second half of the chapter.

In today's class we went through the 2007 trigonometry question which involved material from the first list above. You have to go back quite a few years to find a question that involves compound angles, angles which are > 90 degrees etc.

Homework for tonight was the 2006 Trig question.

In tomorrow's class we will tackle the second part of the trigonometry course.
In the second period, we'll start revising Algebra.

Arithmetic

If you work through all the part a,b and c type questions you got in class today and practice from previous exam papers, you'll be well prepared for Paper 1 Question 1.

We went over 2 types of percentages problems in class:

a) An item is reduced by 10% and the new price is 1125. What was the original price?

and

b) An amount is invested for 1 year at 6% interest rate and amounts to 5618 at the end of the year. What was the original principle invested?

To answer both questions you need to look at the number you are given and work out what percentage this represents.

In a) the amount 1125 represents 90%, so you can write down
90% = 1125 <- This is the most important step. Now I know 90%, I can work out what 10% is by dividing both sides by 9 10% = 125 100% = 1250 <- 100% is the original amount before the reduction. In b) the amount at the end of the year is the original principle (100%) plus 6% so you can write 106% = 5618 1% = 53 [dividing both sides by 106 to find what 1% is] 100% = 5300 <- 100% is the original principle that was invested.

In both cases the first step is key .. look at the number you are given and work out what percentage it represents. Then use the equation technique of multiplying or dividing on both sides of the equals by the same amount to find the percentage you are asked for.

Here is the 1-page overview of arithmetic.


Note that the formula for compound interest can be expressed in two ways.
You can say

1 + (Rate/100)

or

(100 + Rate)/100

The key point is that if the rate is, say, 19%, then you will end up with
Amount = Principle(1.19)^Years

(Sorry about formatting , ^ means raised to a power)