IB Calculus Tutors-Part 1 (Calculus)

In my IB Calculus Tutors series, I will explain different topics taught at HL and SL levels of IB Calculus. Here is the first one.

IB Calculus Tutorial

If we want to understand the importance of Calculus in the IB Diploma Programme, We should have a look at the number of teaching hours recommended for it. It’s 40 hours in SL(out of 150 total hours) and 48 (out of 240 total hours) hours in HL. This makes calculus the most important topic for IB Calculus Tutors as well as for IB students.

What is Calculus-

Calculus is an ancient Latin word. It means ‘small stones’ used for counting. In every branch of Mathematics, we study of something specific, like in Geometry, we study about shapes, in Algebra, we study about arithmetic operations, in coordinate, we study about locating a point. In calculus, we do the mathematical study of continuous change. It mainly has two branches-

►Differential Calculus- Mainly about rates of change and slope of curves

►Integral Calculus-Mainly about areas under the curves, areas between the curves and accumulation of quantities.

These two branches are related to each other by the fundamental theorem of calculus.


Approach to teach calculus- Calculus is a topic which is newer to almost all the students at both SL and HL level and it often takes students a bit more difficult to understand calculus. IB Calculus tutors should keep this thing in mind.
There are many topics in HL which are not there in SL syllabus but we can start both levels in a similar way. Initially, we should start with basic differentiation or the derivative.

Definition of differentiation– The rate of change of any quantity with respect to any other quantity has great importance. The rate of change of any quantity ‘y’ with respect to any other quantity ‘x’, is called the derivative of ‘y’ with respect to ‘x’.

The derivative of a function y=f(x) is the instantaneous rate of change of y or f(x) with respect to the change in the independent variable x.

Derivative of f(x)= {\lim }\limits_{h \to 0} \frac{{change{\rm{ }}in{\rm{ }}x}}{{change{\rm{ }}in{\rm{ }}y}}

As x changes from x to x+h y changes from  f(x) to f(x+h) hence

Derivative of y or     f(x) =  {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}

Process of finding a derivative is called differentiation

we can denote derivative in many ways


All these are different ways to represent derivative

Different Explanations of Differentiation-

Differentiation is explained as a measure of the rate of change in the given function


Differentiation is explained as the velocity of a moving object. this explanation is actually an extension of the first


If distance moved by an object is given by an equation


the velocity of that object at time “t” = \frac{{ds}}{{dt}}

Differentiation is explained as the slope of a tangent line

If equation of a curve is given by y=f(x) then slope the tangent at a point (x,y) =f'(x) or

\frac{{dy}}{{dx}}    at the given point (x,y)

Problem Solving Using Differentiation-

Problems based on differentiation can be solved by two different ways

1. Using the definition of Derivatives (Ab Initio Method)- This method is also known as differentiation by first principle or differentiation by basic rules. This method actually means using the formula mentioned below.

                                        \frac{{dy}}{{dx}} = f'(x) = {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}

there are some results based on special series which are frequently used in this method of differentiation

({\bf{i}})\;\;{{\bf{e}}^{\bf{x}}}\; = {\bf{1}} + \frac{{\bf{x}}}{{1!}} + \frac{{{{\bf{x}}^{\bf{3}}}}}{{3!}} + \frac{{{{\bf{x}}^{\bf{4}}}}}{{4!}} \ldots \ldots \infty


(ii)  ax=1+(xloga)/1!+ (xloga)2/2!+ (xloga)3/3!+ (xloga)4/4!+……….where a > 0


(iii)   ln(1-x)=x-x2/2+x3/3-x4/4……….    where -1 < x  1


(iv)  ln(1-x)=-x-x2/2-x3/3-x4/4……….     where  -1 x < 1


(v )  \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}}.......


(vi) \cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^6}}}{{6!}}.......


(v)  \tan x = x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{5!}} - ..........


(vi)  {\sin ^{ - 1}}x = x + \frac{{{1^2}.x{}^3}}{{3!}} + \frac{{{1^2}.3{}^2x{}^5}}{{5!}} + \frac{{{1^2}{{.3}^2}{{.5}^2}x{}^3}}{{7!}}..........

(vii{\tan ^{ - 1}}x = x - \frac{{x{}^3}}{3} + \frac{{x{}^5}}{5}..........


(viii) {(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{{2!}}{x^2} + \frac{{n(n - 1)(n - 2)}}{{3!}}{x^3}.......


Example-1 Find the derivative of the following function using the definition of the derivative.     



First of all, we shall write down the definition of derivatives

now we should rationalize the numerator of this value by multiplying and dividing by conjugate value

now we can simply cancel “h” and put the limit



 Find the derivative of the following function using the definition of the derivative.

Answer:   First of all, we shall write down the definition of derivatives

Now, we can simplify the whole using algebra


Here is a Pdf of 10 solved questions on differentiation by the first principle

 10 solved questions.pdf

In the next post in IB Calculus Tutors series, I will discuss how to solve problems using differentiation formulas and also the geometrical meaning of differentiation

you can go to the next post and download pdf of calculus questions  by clicking here