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The Chain Rule

October 26, 2010 in General, Grinds, Leaving Cert, MA 1008, MATH6037, MS 2001 | 1 comment

MATH6037 please skip to the end of this entry.

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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Week 5 (Non-Examinable for Test 1…)

October 20, 2010 in MS 2001 | Leave a comment

…but certainly for Test 2!

On Monday we wrote down the definition for a function to be continuous on a closed interval $I\subset \mathbb{R}$ . This is to account for limits such as

$\lim_{x\rightarrow 0}\sqrt{x}$

i.e. in this case the left limit does not exist as the function is not defined for $x<0$ . We wrote down the Intermediate Value Theorem.

On Wednesday we looked at the Intermediate value theorem again. For a proof see http://en.wikipedia.org/wiki/Intermediate_value_theorem We discussed how it implies that, on a closed interval, a continuous function is bounded. If the function is unbounded then it is not continuous. We showed how the Intermediate Value Theorem can estimate the location of roots of functions. Finally we showed with an example the restriction (in the theorem) to closed intervals is necessary.

We then motivated differentiation: why do we do it? We derived the formula for the derivative of a function, defined a differentiable function (at a point) and showed that, as expected, the derivative of a line $y=mx+c$ is just $m$ . This also implies that a constant function has derivative zero. We did one more example (a quadratic). Finally we said that in some sense a differentiable function must be somewhat ‘nicer than’ or ‘as nice’ as a continuous function – as all differentiable functions are continuous. We will prove this on 1/11/10.

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. 9 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise2.pdf

Q. 1,2 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf

Past Exam Papers

Q. 4(b), 4(a) from http://booleweb.ucc.ie/ExamPapers/exams2008/Maths_Stds/MS2001Sum08.pdf

Q. 3(b) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001.pdf

Q. 3(a), 5(a) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/ms2001aut.pdf

Q. 6(a) from http://booleweb.ucc.ie/ExamPapers/exams2002/Maths_Stds/ms2001.pdf

Q. 4(b), 6(a) from http://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf

MS 2001: Week 5 (For Test 1)

October 20, 2010 in MS 2001 | 2 comments

On Monday we again wrote down the definition of a continuous function (at a point). We showed that if we combine continuous functions in ‘nice’ ways they stay continuous. We showed as a corollary that polynomials are continuous everywhere and we stated without proof that $\sin x,\,\cos x$ are also continuous. We defined the composition of two functions and said that, under composition, continuous functions remain continuous (proof on this page). We did a few examples – showing how a function can fail to be continuous by not being defined, by not having a limit, or by the limit not being equal to the value of a function at a point. Finally we said if a function had a discontinuity at a point, if a redefinition of the function at that point could make the function continuous, then the discontinuity is called removable, otherwise it is called essential.

In the tutorial we went through the Sample test. We also did from Ex.Sh.1 Q 7. (vi). Someone asked about Q. 8 (iii) – we didn’t have time to finish it in class but I have the solution on this page.

Finally, for Q. 3 of the test, you need to know the following definitions:

even, odd, increasing, decreasing, quadratic, roots, polynomial, rational function, absolute value, limit, one-sided limit, continuous at a point, continuous, composition

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 (ix), 6, 7, 8 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise2.pdf

Past Exam Papers

Q. 3(b) from http://booleweb.ucc.ie/ExamPapers/Exams2008/MathsStds/MS2001a08.pdf

Q. 1(b) from http://booleweb.ucc.ie/ExamPapers/exams2007/Maths_Stds/MS2001Sum2007.pdf

Q. 3(a) from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/MS2001Sum06.pdf

Q. 3(a) from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/Autumn/ms2001Aut.pdf

Q. 2(b), 3(a) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001.pdf

Q. 3(b) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001Aut05.pdf

Q. 1(b), 3(b) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/MS2001aut.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/MS2001.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2001/Maths_studies/MS2001Summer01.pdf

Q. 2(c) from http://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf

From the Class

1. Recast the definition of a continuous function (at a point) in terms of $\varepsilon$ – $\delta$ .

2. Prove Proposition 3.3.2 using the Calculus of Limits

MS 2001: Tutorial Question

October 19, 2010 in MS 2001 | Leave a comment

Find positive numbers $M$ , $N$ such that $\forall x\in[-1,1]$

$M\leq\left|\frac{2x+7}{5-3x}\right|\leq N$

[Comments in italics]

This problem requires the following fact. For $a,b,c,d\in\mathbb{R}$ , if $0\leq a< b$ and $0\leq c< d$ then we may divide the smaller by the larger and the larger by the smaller to preserve the inequality, i.e.

$\frac{a}{d}\leq \frac{b}{c}$

Now

$\left|\frac{2x+7}{5-3x}\right|=\frac{|2x+7|}{|5-3x|}$

Now, for the upper bound, by the triangle inequality,

$|2x+7|\leq |2x|+7=2|x|+7\leq 9$

As the maximum of $|x|$ as $x\in[-1,1]$ is 1; i.e. for $x\in[-1,1], |x|\leq 1$ . We also used $|xy|=|x||y|$

By the reverse triangle inequality,

$|5-3x|\geq\left||5|-|3x|\right|=\left|5-3|x|\right|$

For $x\in[-1,1]$ , $5-3|x|$ is positive [if $|x|\leq 1$ , then $3|x|\leq 3$ and so $0\leq 3-3|x|\leq 5-3|x|$ ] so

$\left|5-3|x|\right|=5-3|x|$

$\Rightarrow |5-3x|\geq 5-3|x|\geq 2$

We have already seen $3-3|x|\geq 0$ ; add $2$ to both sides.

So we have $0\leq|2x+7|\leq9$ and $0\leq 2\leq|5-3x|$ hence

$\frac{|2x+7|}{|5-3x|}\leq \frac{9}{2}$

Now, for the lower bound, by the reverse triangle inequality:

$|2x+7|=|2x-(-7)|\geq \left||2x|-|-7|\right|=\left|2|x|-7\right|$

For $x\in[-1,1]$ , $2|x|-7$ is negative [if $|x|\leq 1$ , then $2|x|\leq 2$ and so $2|x|-7\leq -5$ ] so

$\left|2|x|-7\right|=7-2|x|$

$\Rightarrow |2x+7|\geq 7-2|x|\geq 2$

We have already seen $2-2|x|\geq 0$ ; add $5$ to both sides.

Now by similar arguments to above:

$|5-3x|\leq |5|+|-3x|=5+3|x|\leq 8$

So we have $0\leq 5\leq|2x+7|$ and $0\leq|5-3x|\leq8$ hence

$\frac{5}{8}\leq \frac{|2x+7|}{|5-3x|}$

Putting these together we get $M=5/8$ and $N=9/5$ :

$\frac{5}{8}\leq\left|\frac{2x+7}{5-3x}\right|\leq \frac{9}{5}$

MS 2001: Sample Test

October 18, 2010 in MS 2001 | Leave a comment

Ye have a test next Wednesday 27/10/10. Please find attached a Sample

I will not be providing solutions to this sample. If you want solutions please attend the tutorial and ask me to do a question from the sample test.

Everything up to but not including section 3.4 (which we will reach today Monday 18/10, or on Wednesday 20/10) ( i.e. everything up to exercise 2.31 in Wills’ notes MS2001. ) will be examinable.

Question 1 will be taken from the exercise sheets, Question 2 from a past exam paper, and Question 3 will be on definitions (presented in class and also in Wills’ notes.)

The Composition of Continuous Functions is Continuous

October 18, 2010 in MA 1008, MS 2001 | 1 comment

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,

$\Box$

MS 2001: Week 4

October 13, 2010 in MS 2001 | 2 comments

On Monday we wrote down the Calculus of Limits. We stated that parts (iii) and (iv) have nasty proofs and left them on the webpage. We proved parts (i) and (ii) and showed in part (v) that if the limit exists, it must have that form. We showed that for any polynomial $p(x)$ , that $\lim_{x\rightarrow a}=p(a)$ . Finally we proved the special limit:

$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$

In the tutorial we ran out of questions fairly early, so I threw the exercises up on the projector and allowed ye work away. Ye put up your hand if ye wanted assistence.

On Wednesday we considered infinite limits – when does a function tend to infinity and what does a function do as $x$ tends to infinity. Also we gave the crucial definition of a continuous function.

Finally I stated that the Sample Test will be up before or on Wednesday 20 – the test is on 27/10/10. Question 1 will be from the exercises, Question 2 from past exam papers, and Question 3 will be about definitions.

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.10 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise1.pdf

Q. 1\(ix), 2,3,4,5 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise2.pdf

More exercise sheets

Section 2 from Problems

Past Exam Papers

Q. 1(b), 3(a) from http://booleweb.ucc.ie/ExamPapers/exams2010/MathsStds/MS2001Sum2010.pdf

Q . 1(b), 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2009/MathsStds/MS2001s09.pdf

Q. 1(b) from http://booleweb.ucc.ie/ExamPapers/exams2009/MathsStds/Autumn/MS2001A09.pdf

Q. 1(b), 3(a) from http://booleweb.ucc.ie/ExamPapers/exams2008/Maths_Stds/MS2001Sum08.pdf

Q. 1(b), 2(a)(ii-iii), 3(a) from http://booleweb.ucc.ie/ExamPapers/Exams2008/MathsStds/MS2001a08.pdf

Q. 1(a) from http://booleweb.ucc.ie/ExamPapers/exams2007/Maths_Stds/MS2001Sum2007.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/MS2001Sum06.pdf

Q. 2(b) from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/Autumn/ms2001Aut.pdf

Q. 2(b), 3(a) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001.pdf

Q. 2(b), 3(a) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001Aut05.pdf

Q. 1(b), 3(b) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 1(b)from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/MS2001aut.pdf

Q. 1(b) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/MS2001.pdf

Q. 1(b), 2(a) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/ms2001aut.pdf

Q. 1(b) from http://booleweb.ucc.ie/ExamPapers/exams2002/Maths_Stds/ms2001.pdf

Q. 1(a), 2(a), 3 from http://booleweb.ucc.ie/ExamPapers/exams2001/Maths_studies/MS2001Summer01.pdf

Q. 3 from http://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf

From the Class

1. Prove the following proposition:

Suppose that $f,g$ are functions for which $f(x)=g(x)$ for all $x \neq a$ . If $\lim_{x\rightarrow a}f(x)$ exists then so does $\lim_{x\rightarrow a}g(x)$ and moreover they are equal.

2. Investigate, for $n$ an odd natural number,

$\lim_{x\rightarrow -\infty}x^n$

3. Investigate

$\lim_{x\rightarrow -2}\frac{x-3}{x^2+3x+2}$

MS 2001: Announcement 2 Update

October 6, 2010 in MS 2001 | Leave a comment

Those who had issues with clashing tutorials, please see your student email inbox.

Calculus of Limits: Product and Quotient Rules

October 6, 2010 in General, Grinds, Leaving Cert, MA 1008, MS 2001 | 4 comments

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

$\lim_{x\rightarrow a} (f(x)+g(x))=p+q$ .
If $k\in\mathbb{R}$ , $\lim_{x\rightarrow a} kf(x)=kp$ .
$\lim_{x\rightarrow a} (f(x)g(x))=pq$ .
If $q\neq0$ , $\lim_{x\rightarrow a} (f(x)/g(x))=p/q$ .
If $n\in\mathbb{N}$ , and $p>0$ then $\lim_{x\rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{p}$ .
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MS 2001: Announcement 3 (Test)

October 6, 2010 in MS 2001 | Leave a comment

The first in-class test will take place on 27 October 2010. Any material presented in class, up to and including 20 October is examinable. The test is worth 12.5% of your continuous assesment mark for MS 2001. A sample test shall be posted here on 20 October 2010.

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Category Archive

The Chain Rule

Week 5 (Non-Examinable for Test 1…)

MS 2001: Week 5 (For Test 1)

MS 2001: Tutorial Question

MS 2001: Sample Test

The Composition of Continuous Functions is Continuous

MS 2001: Week 4

MS 2001: Announcement 2 Update

Calculus of Limits: Product and Quotient Rules

MS 2001: Announcement 3 (Test)

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