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JEE Main: Differentiation and How To Calculate ‘e’

JEE Main: Differentiation and How To Calculate ‘e’

Edited By Ramraj Saini | Updated on Apr 19, 2024 07:49 PM IST | #JEE Main

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?

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).

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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…).

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Questions related to JEE Main

Have a question related to JEE Main ?
my son got 96.9 in Jee mains with 47K rank. can you let me know the probable NITs and the course types.

Admissions to any universities or colleges are given based on Rank. As you mentioned total rank is 47000. It is possible to get admission into some good NITs.

To know more about NITs cutoff for 2024, click on the below link.

https://engineering.careers360.com/articles/jee-main-cutoff-for-top-nits


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Rishitha 30 April,2024
sir ,i have done my 12th in 2021 with pcb .now i wants to get into iisc banglore. so decided to repeat 12 with pcmb and give jee and neet in 2025 ,is am eligible for get into iisc banglore.

Hello khushabu

Whether You will be eligible to get a seat in iisc depends upon your jee mains score and cut off iisc jee in 2025.

For general category , the candidate must get around 350 rank in jee mains

Around 17000 rank for ews and obc  category

Around 20000 for obc and ews ( female )

Now a days there has been introduced a New method , iiser aptitude test

The eligibility criteria for getting a seat in iisc , the candidate must obtain minimum 60% marks in class xii.

In physics,  chemistry and mathematics.

10 % seats are reserved for women.

Read Complete Answer
Manisha 29 April,2024
i have got 3.5 lakh rank in jee mains which college will I get in cse

At this score you can go for private colleges, because govt colleges will be tough at this rank.

Even you can try for State counselling where private colleges do comes under that counselling.

To know in detail you can check your chances through our JEE Main College Predictor Tool.

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Hritik.maheshwari 30 April,2024
my percentile is 48 in jee mains and I'm obc candidate I'm eligible or not for jee advance exam

Hello Student ,

As the qualifying  percentile to become eligible for JEE Advanced this year is 79.6757881 for OBC category so according to this you are not eligible.
To get a detailed information about the branches you will get on the basis of JEE Main rank use JEE College Predictor by Careers360.
Link for College predictor - https://engineering.careers360.com/jee-main-college-predictor

Thanks

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ruchika.chauhan 29 April,2024
sir I got 98.96 percentile in jee mains .can I get seat in iiitb.

Hello aspirant,

If we look at previous years trends, then having a percentile of 98.6 in JEE Main means you won't be able to get admission to IIIT Bangalore for Integrated MTech in either CSE or ECE. The cutoff of IIT Bangalore go very high and you must me among the top students to get admission there.

Thank you

Hope this information helps you.

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Sajal Trivedi 29 April,2024
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