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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).

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…).

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Questions related to JEE Main

Have a question related to JEE Main ?

Scoring a 92 percentile in the JEE Main is a commendable achievement. However, securing a computer science and engineering seat in colleges in Coimbatore would depend on several factors.

Possible colleges in Coimbatore are:

  1. PSG College of Technology
  2. Coimbatore institute of technology
  3. Kumaraguru College of Technology
  4. Amrita Vishwa Vidyapeetham


Hello aspirant,

Unfortunately it is impossible for you to get any good college ( whether government or private) with 64 percentile in jee mains. It is one of the toughest exam of India and its cutoff is also very high. I suggest you to take a drop and give your best for next attempt.

Thank you

Hope this information helps you.

Hello,


Yes, you can repeat your 12th grade even after passing, including on the Telangana State Board of Intermediate Education (TSBIE). There are several reasons why students choose to repeat a year, such as aiming for higher marks, switching to a different stream, or improving their subject knowledge for competitive exams.


Hope this helps,


Thank you

Yes, you can write the NIOS exam a year after passing 12th standard. There is no restriction on appearing for NIOS based on the gap since your previous 12th board exams.  NIOS qualification is perfectly acceptable for JEE Mains. You just need to meet the eligibility criteria, which includes passing 10+2 with Physics, Chemistry, and Mathematics. JEE Mains considers the year of your first attempt at 12th board exams (NIOS or any other board) to determine eligibility. You can only appear for JEE Mains for three consecutive years after that first attempt year.

Similar to JEE Mains, BITSAT also accepts NIOS qualifications. BITSAT eligibility is based on the year you passed or appeared for your 12th board exams (current year and previous year). So, you can appear for BITSAT the year you take the NIOS exam and the following year as well.

https://www.careers360.com/exams/bitsat

I hope it helps!


Hello,


With a 69 percentile in JEE Main, it is highly unlikely for your son to secure admission in IIT Bombay. Admission to IITs, including IIT Bombay, is based on the JEE Advanced exam, for which qualification through JEE Main is required. Generally, to qualify for JEE Advanced, candidates need to be in the top 2,50,000 ranks in JEE Main, which typically requires a higher percentile than 69.


Hope this helps,


Thank you

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 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C.  If Na2SO4 is 81.5% ionised, the value of x (Kf for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Option: 1  15 g
Option: 2  25 g
Option: 3  45 g
Option: 4  65 g  
 

 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.  If pKb of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75
Option: 3 8.25
Option: 4 9.25
 

CH_3-CH=CH-CH_3+Br_2\overset{CCl_4}{\rightarrow}A

What is A?

Option: 1

CH_3-CH(Br)-CH_2-CH_3


Option: 2

CH_3-CH(Br)-CH(Br)-CH_3


Option: 3

CH_3-CH_2-CH_2-CH_2Br


Option: 4

None


\mathrm{NaNO_{3}} when heated gives a white solid A and two gases B and C. B and C are two important atmospheric gases. What is A, B and C ?

Option: 1

\mathrm{A}: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 2

A: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 3

A: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


Option: 4

\mathrm{A}: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


C_1+2 C_2+3 C_3+\ldots .n C_n=

Option: 1

2^n


Option: 2

\text { n. } 2^n


Option: 3

\text { n. } 2^{n-1}


Option: 4

n \cdot 2^{n+1}


 

A capacitor is made of two square plates each of side 'a' making a very small angle \alpha between them, as shown in the figure. The capacitance will be close to : 
Option: 1 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{4 d } \right )

Option: 2 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )

Option: 3 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )

Option: 4 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )
 

 Among the following compounds, the increasing order of their basic strength is
Option: 1  (I) < (II) < (IV) < (III)
Option: 2  (I) < (II) < (III) < (IV)
Option: 3  (II) < (I) < (IV) < (III)
Option: 4  (II) < (I) < (III) < (IV)
 

 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.  If during  this process the relation of pressure P and volume V is given by PVn=constant,  then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
Option: 1  n=\frac{C_{p}}{C_{v}}


Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


Option: 3 n=\frac{C_{p}-C}{C-C_{v}}

Option: 4  n=\frac{C-C_{v}}{C-C_{p}}
 

As shown in the figure, a battery of emf \epsilon is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is : 


Option: 1 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
Option: 2 \frac{\epsilon L }{R^{2}}


Option: 3 \frac{\epsilon R }{eL^{2}}

Option: 4 \frac{\epsilon L }{eR^{2}}
 

As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed  of the particle at B is x m/s. (Take g = 10 m/s2 ) The value of 'x' to the nearest is ___________.
Option: 1 10
Option: 2 20
Option: 3 40
Option: 4 15

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