IB Math Tutors series- Calculus

Our  Ib Math tutors understand that calculus is one of the most important topics. 40 teaching hours are given for Sl calculus(27% of total time) and 48 hours are given for HL calculus(20% of total time) and that is maximum in IB math. After understanding the importance of calculus we have written many posts on this topic. A few of them are:

How to solve limit problems

Continuity of Functions

Applications of derivatives

Increasing and decreasing functions

Maxima and Minima

definite integration

Indefinite integration

Applications of Integration

Read more

How to get best results from your online ib tuition

IB Elite Tutor works with its motive to “Develop a passion for learning”. We are a premier tutoring institute that has been founded in 2011 to provide Online IB tuitions and IB tuitions in Delhi,Gurgaon,Noida,Banglore,Hyderabad,Mumbai, and Chennai to the students of International BaccalaureateIB MYP, IB IA, IB DP (all Groups-HL, SL), Cambridge A&AS Level, O-level,IGCSE/GCSE, SAT/ACT (all subjects) in this competitive time with the most experienced, trained and qualified online IB tuitions experts. Our online ib tuitions are carried in the simplest way.

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Binomial Theorem

The formula by which any positive integral power of a  binomial expression can be expanded in the form of a series is known as  Binomial  Theorem. If   x, y ∈ R and n∈N,  then

(x + y)n = nC0 xn + nC1 xn-1 y + nC2 xn-2y2 + ….. + nCrxn-r yr + ….. + nCnyn =nCr xn – r yr

This theorem can be proved by Induction method.

<img src="binomial theorem.png" alt="binomial theorem">


(i)  The number of terms in the expansion is (n + 1) i.e. one or more than the index.

(ii) The sum of the indices of x & y  in each term is n.

(iii) The binomial coefficients of the terms nC0, nC1……..  equidistant from the beginning and the end are equal. Read more

Calculus (Part-2)-geometric explanation of differentiation

geometric explanation of differentiation

Geometric explanation of differentiation-  If we find the derivative of function f(x) at x = x0 then it is equal to the slope of the tangent  to the graph of given function f(x) at the given point [(x0, f(x0))].<img src="derivative.png" alt="derivative">

But what is a tangent line?

It is not merely a simple line that joins the graph of given function at one point.

It is actually the limit of the secant lines joining points P = [(x0, f(x0)] and Q on the graph of f(x) as Q moves very much close to P.

The tangent line contacts the graph of given function at the given point [(x0, f(x0)] the slope of the tangent line matches the direction of the graph at that point. The tangent line is the straight line that best approximates the graph at that point.

<img src="derivative and slope.png" alt="derivative and slope">


As we are given the graph of given function, we can draw the tangent to this graph easily. Still, we’ll like to make calculations involving the tangent line and so will require a calculative method to explore the tangent line.


We can easily calculate the equation of the tangent line by using the slope-point form of the line. We slope of a line is m and its passing through a point (x0,y0) then its equation will be

                                                              y − y0 = m(x − x0)


So now we have the formula for the equation of the tangent line. It’s clear that to get an actual equation for the tangent line, we should know the exact coordinates of point P. If we have the value of x0 with us we calculate y0 as

                                                                      y = f(x0)

The second thing we must know is the slope of line

                                                                      m = f’(x0)

Which we call the derivative of given function f(x).



The derivative f’(x0) of given function f at x0 is equal to the slope of the tangent line to

                                        y = f(x) at the point P = (x0, f(x0).

Differentiation Using Formulas- We can use derivatives of different types of functions to solve our problems :


{\rm{(i) D (}}{{\rm{x}}^{\rm{n}}}{\rm{) = n}}{\rm{.}}{{\rm{x}}^{{\rm{n - 1}}}}{\rm{ here n, x}} \in R        \;\left( {ii} \right){\rm{ }}\;D{\rm{ }}\left( {{e^x}} \right){\rm{ }} = {\rm{ }}{e^x}

{\rm{(iii) D(}}{{\rm{a}}^{\rm{x}}}{\rm{) = }}{{\rm{a}}^{\rm{x}}}{\rm{lo}}{{\rm{g}}_e}{\rm{a a > 0}}                {\rm{(iv) D (ln x) = }}\frac{1}{x}

{\rm{(v) D(lo}}{{\rm{g}}_{\rm{a}}}{\rm{x) = lo}}{{\rm{g}}_{\rm{a}}}{\rm{e }}                      {({\rm{vi}}){\rm{D}}({\rm{sinx}}){\rm{ = cosx}}}

{({\rm{vii}}){\rm{D}}({\rm{cosx}}){\rm{ = - sinx}}}                     {({\rm{viii}}){\rm{D(tanx) = se}}{{\rm{c}}^2}x}

(ix)  D (secx) = secx . tanx              (x)  D (cosecx) = – cosecx . cotx   

{\rm{(xi) D (cotx) = - cose}}{{\rm{c}}^2}x              (xii) D (constant) = 0 where D = {\frac{d}{{dx}}}

These formulas are result of differetiation by first principle

Inverse Functions And Their Derivatives :


{\rm{(i) }}\frac{d}{{dx}}({{{\rm Sin}\nolimits} ^{ - 1}}x) = \frac{1}{{\sqrt {{1^2} - {x^2}} }}                 {\rm{(ii) }}\frac{d}{{dx}}({\cos ^{ - 1}}x) = \frac{{ - 1}}{{\sqrt {{1^2} - {x^2}} }}

{\rm{(iii) }}\frac{d}{{dx}}({{{\rm Sec}\nolimits} ^{ - 1}}x) = \frac{1}{{x\sqrt {{x^2} - 1} }}              {\rm{(iv) }}\frac{d}{{dx}}({{{\rm cosec}\nolimits} ^{ - 1}}x) = \frac{{ - 1}}{{x\sqrt {{x^2} - 1} }}

{\rm{(v) }}\frac{d}{{dx}}(Co{t^{ - 1}}x) = \frac{{ - 1}}{{1 + {x^2}}}                  {\rm{(vi) }}\frac{d}{{dx}}({\tan ^{ - 1}}x) = \frac{1}{{1 + {x^2}}}

Theorems On Derivatives:  If u and v are derivable function of x, then,


(i)   \begin{array}{l} {\rm{ }}\frac{d}{{dx}}(u \pm v) = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}\\ \\ \end{array}                    (ii)   \frac{d}{{dx}}k[f(x)] = k\frac{d}{{dx}}f(x)    where K is any constant


(iii)  {\rm{ }}\frac{d}{{dx}}(u.v) = v\frac{{du}}{{dx}} + u\frac{{dv}}{{dx}}    known as  “ Product  Rule ”      (iv)   {\rm{ }}\frac{d}{{dx}}(\frac{u}{v}) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}

known as  “Quotient  Rule ”


(v) If  y = f(u)  &  u = g(x)  then  \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}    “Chain  Rule ”


Logarithmic  Differentiation:  To find the derivative of  : (i) a function which is the product or quotient of a number of functions OR
(ii) a function of the form  {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} where f & g  are both differentiable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic  Differentiation.


Implicit  Differentiation:  (i) In order to find dy/dx,  in the case of implicit functions, we differentiate each term w.r.t.  x  regarding y as a function of x & then collect terms in dy/dx together on one side to finally find dy/dx.

(ii) In answers of dy/dx in the case of implicit functions, both x & y are present.

Parametric  Differentiation: If  y = f(q)  &  x = g(q)  where q  is  a parameter, then

                                                                         \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }}

Derivative Of A Function w.r.t. Another Function-: Let  y = f(x) & z = g(x)

\frac{{dy}}{{dx}} = \frac{{f'(x)}}{{g'(x)}}

Derivatives Of Order Two & Three: Let a function y = f(x) be defined on  an open interval (a, b).  It’s derivative, if it exists on(a, b)  is  a certain function f'(x) [or (dy/dx) or y’ ] &  is  called  the  first derivative of y  w.r.t. x. If it happens that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of y  w. r. t.  x  &  is denoted by f”(x) or (d2y/dx2) or  y”. Similarly, the 3rd order derivative of y w. r. t.  X, if it exists, is defined by    It is also denoted by f”'(x) or y”’.

All  maths tutors suggest solving a fair amount of questions based on the geometric explanation of differentiation


Example-Find the tangent line to the following function at z=3 for the given function


We can find the derivative of the given function using basic differentiation as discussed in the previous post            

We are already given that z=3 so

Equation of tangent line is

                                                    y − y0 = m(x − x0)      here y0=R(3)=√7

Putting these values we get equation of tangent line


So that’s all for the geometric explanation of differentiation, I will discuss Application of Derivatives in my next post


Example-  Differentiate the following function    


Ans: We can apply quotient rule in this questions


You can try these questions in the meantime

 28 – Differentiation (09-10-15) with answers.pdf



<img src="demo.png" alt="demo">



IB Mathematics Tutors-Part 1 (Calculus)

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

<img src="ib mathematics tutors.jpg" alt="ib mathematics tutors">

If we want to understand the importance of Calculus in 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 Mathematics 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- Read more

How to find sum and product of zeros of equations

In my previous post on IB Mathematics, I have discussed how to solve a quadratic polynomial using Quadratic formula. Here I will tell you about different relationships based on sum and product of quadratic polynomial, cubic polynomial and bi-quadratic polynomials.

<img src="ib mathematics.png" alt="ib mathematics">


ax2 + bx + c = 0

                                         Sum of the roots = −b/a

                                          Product of the roots = c/a

If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula

x2 − (sum of the roots)x + (product of the roots) = 0 Read more