[H2_Expository] Derivative of the Arcsecant function

Introduction

     This article explains how to differentiate, i.e. find the derivative of, the arcsecant function, which is seldom discussed in class.  If you look at the various syllabus outlines, sometimes they do not explicitly mention or imply the arcsecant, neither do they explicitly exclude this.  The important thing is: we should be able to figure it out from our basic knowledge, which is the whole point of mathematics.  The people who talk about and focus on mathematics syllabus content as if it is the only or most important thing are missing the point of mathematics education, and there are a plenty of these idiots around.  Do not follow them.

     The arcsecant, written as arcsec or sec^-1, is also known as the inverse secant function.  So arcsecant means that, given the value of a secant function, you want “the” angle whose secant is that given value.  The problem is there are many possible values.  Look at the graph below.

Defining the Inverse of the secant properly

     On the graph, a horizontal line can pass through (infinitely) many points.  Like any periodic trigonometric function, the secant function is not a one-to-one (a.k.a. “injective” or  “one-one”) function.  As such, it is not invertible.  However, we can restrict the function so that its domain is  [0, p] \ { ^p/₂ }.  This is highlighted in yellow on the graph.

     Remember that sec x = ¹/_{cos x}.  Basically we follow the principal values  [0, p]  of the arccosine function   except  ^p/₂  where the cosine is zero and its reciprocal the secant is undefined.       With this restriction on the domain, we get a one-one secant function, with range (-¥, -1] È [1, +¥).  We can now define the inverse function, and its graph is obtained by reflecting the above graph along the mirror line  y = x.  We get this:-

     Observe that in the yellow regions in both graphs, the gradients at the points are non-negative.  We have chosen the domain of the secant function, which is the same as the range of the arcsecant function, such that the derivatives will be non-negative.

Deriving the Derivative (refer to the“Onion” Method for differentiation)

Remarks

     Do not confuse  arccos y   with  (cos y)^-1.  They are not mean the same thing.

     In case you are wondering, the prefix “arc-” means the angle, which (if you use radians) is literally the same as the arc-length when the radius equals to  1.  In symbols, s = rq  with r = 1 means  s = q.  Although the term can be used with degrees or other units, when doing advanced mathematics like calculus, we would usually be using radians anyway. 

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* International Baccalaureate (IB) HL Mathematics

* Advanced Placement (AP) Calculus AB & BC

* University / College calculus

* other syllabuses that involve differentiation

* any learner interested in calculus

JCCDQBHWH_FN021(b) Range of Composite Functions

[original source unknown]

Introduction

     This question is from the usual book which did not credit the source.  It comes from some unknown junior college in an unknown year.  It is a challenging question because most students are poor at finding the range of composite functions.  Furthermore, this question has a little twist: you are given the range of the composite function, but you are required to solve for something.  So you need to, in a way, work backwards and/or use inequalities (another weak point for many students).

     Students are reminded of the right-to-left convention for functions in JC as well as GCE ‘A’ level examinations.  This means in ‘fg’ the  g  is done first before the  f.  There are some university professors who use a left-to-right convention, but here we do not.  So take note.

     Again metacogntion and heuristics are very important and I will illustrate their use.

Stage 1:  Understanding the Problem

What is this (part of a) question about?

range of composite functions, solving for unknown

What is given in the question?

The range of  fg.

What is the question asking for?

The value of  k  that leads to the given range.

Stage 2:  Planning the strategy

What heuristics do you think can be used for this question?

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

Can you recall the definition of the range of a function?  The range of a composite function?

Stage 3:  Execution

Any observations that can make your job simpler?

     Yes.  Observe that  g(x)  is a quadratic with positive  x²  coefficient.  So this is a parabola that looks like a happy smile.  To locate the minimum point, we can complete the square (a technique learnt in secondary school).

     g(x)  =  x² + 2x – 1  =  x² + 2x + 1² – 1² – 1  =  (x + 1) ² – 2

when  x = -1,  g(x) = -2.  The minimum point is (-1,-2).  So the range of  g  is all the

numbers from  -2  upwards.  i.e.  R_g = [-2,` \oo `).

So what now?

     With the two-stage method, suppose now we have

                  x ` \in ` R_g

That means?

                  x > -2

That means?

                  x + k + 1 > -2 + k + 1

Why do you do that?

     I want to slowly manipulate the LHS to get  ln(x + k + 1)  which is  f(x).  Continuing,

                  ln(x + k + 1) > ln(k – 1)

                                f(x) > ln(k – 1)

i.e.                            R_fg = [ln(k – 1), ` \oo `).

Why is there no switching in the inequality sign?

     The slope of the graph of  the natural logarithm is always positive (albeit getting less steep for increasing  x).  So applying  ‘ln’  on both sides does not change the inequality.

What is the clue again?

     We are told that  R_fg = [ln 3, ` \oo `).  Aha!  *epiphany*  *light bulbs flashing*

ln(k – 1) must be equal to ln 3!!!  This can be solved easily!

figure 1 – working forward and backwards

Stage 4:  Evaluation

Is the answer correct?

     Substituting  k = 4,  we see that    ln(x + k + 1) =  ln(x + 5)  and with  x > -2,  this will be  > ln 3  as given in the clue.

And why  x > -2?

     This is because g(x) > -2, which we knew  from completing the square.  We treat the  ‘g(x)’  as the  ‘x’  when applying  f,  because this is what  fg(x)  really means.

Stage 5:  Reflection

What did we learn from solving this question?

     We used metacognition to do self-monitor and self-questioning during the 5 stage problem-solving process.

     We used the following heuristics.

· Working forwards (considering the meanings, asking “so what?” “what next?”)

· Setting up equation/inequality

· Working/thinking backwards

· Consider equivalent expressions or rephrasing the problem

     We learned to apply the definition of the range of fg.  There are two possible methods: the one-stage method and the two-stage method.  The latter is usually better.

     In the two-stage method, the range of  fg  is found by first finding the range of  g  and then applying the function  f  to it.  After the first step of finding the inequality for  g(x),  we can simply use  x  in the formula for  f.  How?  We set  x  to be in the range of  g(x) from the previous step,  then slowly manipulate the inequality until the expression for  f(x)  appears.  This will give us the range of  fg.

     From the formula for the range of  fg,  we learned how to make use of the given clue to work backwards to find the unknown k.

     Difficult questions can be tackled by thinking systematically and logically, and using heuristics and metacognition.  Mathematics is hard, but it is fun after you have learned it.  If you have really learned it, you become more powerful because you can use the same technique to solve all kinds of problems in future.

Any of your own reflections?  Please post in the comments below.