Careers360 Logo
ask-icon
share
    JEE Main: Differentiation and How To Calculate ‘e’

    JEE Main: Differentiation and How To Calculate ‘e’

    Ramraj SainiUpdated on 19 Apr 2024, 07:49 PM IST

    We all have come across the number ‘e’ while studying logarithm and calculus. This number is an irrational number like π, and is called Euler’s Number. Its value is 2.71828… and this sequence of digits never ends. So, how can we find this number and what is so special about this number?

    JEE Main: Differentiation and How To Calculate ‘e’
    Exponential-graph(Image:Wikimedia commons)

    A property that you must have studied in differentiation that stands out among all differentiation formulae is

    \frac{d}{dx}e^x=e^x

    So, the differentiation of the function f(x) = ex, is this function itself. This makes this function unique. We will be using this property to calculate the number ‘e’. So, in all the calculations below, we will not use ‘e’ directly.

    Let us start with the differentiation of a general exponential function, f(x) = 2x.

    Using First Principle of Differentiation, we know that the differentiation of a function f(x) is

    f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

    So, differentiation of f(x) = 2x will be

    \\f'(x)=\lim_{h\rightarrow 0}\frac{2^{x+h}-2^x}{h}\\\\f'(x)=\frac{2^x.2^h-2^x}{h}\\\\f'(x)=2^x\lim_{h\rightarrow 0}\frac{2^h-1}{h}.......equation(1)

    Now, when the value of the limit h→0 [(2h-1)/h] is calculated by putting different values of h that are very close to 0, we can see that the value of this expression approaches 0.693147…(Note that we are not directly using the value of this limit as ln(2), as the number ‘e’ and thus ln(x), which is log with the base ‘e’, is not yet known).

    Also Read,

    Let us see what values of [(2h-1)/h] we get by putting different values of h that are close to 0. We can use a calculator to find these values

    When h = 0.001, [(2h-1)/h] = 0.6933…

    When h = 0.0001, [(2h-1)/h] = 0.6931…

    When h = 0.00000001, [(2h-1)/h] = 0.6931…

    We can see that [(2h-1)/h] value approaches 0.6931…and hence

    limit h→0 [(2h-1)/h]= 0.6931…

    From equation (i):

    f'(x) = 2x.(0.6931…)

    So, differentiation of f(x) = 2x is of the form

    f'(x) =  Some constant. f(x)

    Now if we do the same procedure with f(x) = 4x

    \\f'(x)=\lim_{h\rightarrow 0}\frac{4^{x+h}-4^x}{h}\\\\f'(x)=\frac{4^x.4^h-4^x}{h}\\\\f'(x)=4^x\lim_{h\rightarrow 0}\frac{4^h-1}{h}.......equation(2)

    When h = 0.001, [(4h-1)/h] = 1.3872…..

    When h = 0.0001, [(4h-1)/h] = 1.38631…

    When h = 0.00000001, [(4h-1)/h] = 1.38629…

    So, limit h→0 [(4h-1)/h] = 1.38629…

    And from (ii),

    f'(x) =  4x . (1.38629…)

    So, differentiation of f(x) = 4x is again of the form

    f'(x) =  Some constant. f(x)

    In fact, we can do the same exercise for any positive real number a, and we will find that the differentiation of f(x) = ax equals some constant times ax

    It can also be seen that the value of this constant keeps on increasing as the value of ‘a’ increases. For example

    For a = 2, the constant we calculated was 0.6931…

    For a = 4, the constant we calculated was 1.38629…

    Similarly, for a = 5, the constant can be calculated to be 1.6094…

    For a = 6, the constant is 1.7917…

    So naturally we can ask ourselves the question that can we find a number ‘a’ for which this constant value equals 1, and thus differentiation of ax is 1. ax, meaning that the differentiation of the function ax is this function itself ( = ax)

    After doing multiple hits and trials, this number can be found to be 2.71828. That is why we have the unique property,

    d/dx(ex)= ex

    Euler’s Number also finds applications in fields of mathematics other than calculus. One of the most important applications is in Complex Numbers. You must have come across the relation eiπ = - 1. Imaginary powers of e help us get the values of many trigonometric series which would otherwise be very difficult to prove using only the trigonometric relations. The number is also used in Finance (to calculate compound interest), to explain population growth of humans or microbes, to explain radioactive decay (which in turn is used to tell the age of ancient objects), etc.

    Due to numerous applications, ‘e’ is the second most famous mathematical constant after π. We also celebrate ‘e-day’ on 7 February. This date is chosen as it is written as 2/7 in month/date format and the digits 2,7 represent the first two digits used in the value of ‘e’ (2.71…).

    Articles
    |
    Certifications By Top Providers
    Data Analytics with Python
    Via Indian Institute of Technology Roorkee
    Basic Programming using Python
    Via Indian Institute of Technology Bombay
    Biomedical Nanotechnology
    Via Indian Institute of Technology Roorkee
    Online MCA
    Via Dayananda Sagar University, Bangalore
    Material Science
    Via Indira Gandhi National Open University, New Delhi
    Biomechanics of Joints and Orthopaedic Implants
    Via Indian Institute of Technology Kharagpur
    Explore Top Universities Across Globe

    Questions related to JEE Main

    On Question asked by student community

    Have a question related to JEE Main ?

    Hello Dear Student,


    With a JEE Main CRL of around 3,00,000 and an EWS rank of around 46,000 , getting Computer Science and Engineering (CSE) at Odisha University of Technology and Research is highly unlikely .

    Why?

    • CSE is one of the most competitive branches at OUTR.
    • Previous years' closing

    Hey there,

    Yes, with 62% in Class 12 PCM, you can get direct admission to B.Tech in many AICTE-approved private engineering colleges without appearing for JEE Main. Many private universities and engineering colleges offer admission based on Class 12 marks, institute-level admission processes, or management quota seats. Generally, the minimum

    Hello Dear Student,

    Yes. CBSE allows a student to appear for Improvement Examinations as a private candidate in the succeeding year .

    For JoSAA eligibility , what matters is meeting the Class 12 performance criterion applicable for that admission year. If your improvement marks increase your aggregate and you satisfy

    Hello Dear Student,
    No, with 89.53 percentile in JEE Mains and 79.83 percentile in MHT CET, you will not be able to get Production Engineering at Veermata Jijabai Technological Institute (VJTI).

    You can check, find and access more information here: https://engineering.careers360.com/articles/vjti-cutoff-mht-cet-2026-branch-wise-percentile-analysis-for-vjti-mumbai-admission
    Hope it helps!