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# JEE Main Study Notes for Differential Calculus: Basic Concepts, Formulas, Tips & Tricks Differential Calculus is the easiest part of calculus in JEE Mains Paper and there is no doubt in the fact that the topic is scoring too. The topic is very interesting and consists of 6-7 questions in the paper that totals to 28 marks in the overall marks distribution of the paper. Some of the topics are Differentiation, Tangent and Normal, Maxima and Minima, Differential Equations, Limits, Continuity and Differentiability etc. These topics are quite fascinating but involve concepts which must be understood properly. It is the most scoring part of the IIT JEE Mains Paper. Read JEE Main Mathematics Syllabus

1. Minimum of 6-7 questions will be asked in the JEE Mains Paper from Differential Calculus.
2. The weightage of the marks of Differential Calculus in JEE Mains is 28 marks.
3. Some important topics from Differential Calculus are Differentiation, Limits, Continuity and Differentiability, Differential Equations, Tangent and Normal, Maxima and Minima etc.

### Important Topics from Differential Calculus in JEE Main

The table below provides important topics from Differential Calculus that have consistently appeared in JEE Main along with the number of questions and distribution of marks.

TopicsNumber of QuestionsMarks
Limits, Continuity and Differentiability312
Differential Equations14
Differential Calculus14
Application of Derivatives: Tangent and Normal, Maxima and Minima, Area and Volume etc.24

Below given are some of the detailed topics along with the questions based on Differential Calculus for the IIT JEE Mains point of view.

### Limit of a Function

Suppose f: R → R is defined on the real line and p, L ∈ R. Then, we can say that the limit of a function f is l if

For every real ε > 0, there exists a real δ > 0 such that for all real x,

0 < | x − p | < δ implies | f(x) − L | < ε.

Mathematically, it is represented as

Example: Find the limit of the function f(x) = (x2-6x + 8) / x-4, as x→5.

Solution:

The limit is 3 because f(5) = 3 and this function is continuous at x = 5.

### Continuity of a Function

A function y = f(x) is continuous at point x = a if the following three conditions are satisfied:

1. f(a) is defined.
2. exists (i.e., it is finite)

A function can also be continuous when its graph is a single unbroken curve. This definition of a continuous function is useful when it is possible to draw the graph of a function so that just by the graph the continuity of the function can be judged.

Example: The number of values of x ∈ [0, 2] at which f (x) = ∣x − [1 / 2]∣ + |x − 1|+ tanx is not differentiable is

1. f(a) is defined.
2. exists (i.e., it is finite)
1. 0
2. 1
3. 3
4. None of these

Solution: ∣x − [1 / 2]∣ is continuous everywhere but not differentiable at x = 1 / 2, |x − 1| is continuous everywhere, but not differentiable at x = 1 and tan x is continuous in [0, 2] except at x = π / 2. Hence, f (x) is not differentiable at x = 1 / 2, 1, π / 2.

Example: Which of the following functions have a finite number of points of discontinuity in R ([.] represents the greatest integer function)?

1. tanx
2. x[x]
3. |x| / x
4. sin[πx]

Solution:

f (x) = tanx is discontinuous when x = (2n + 1) π / 2, n ∈ Z

f (x) = x[x] is discontinuous when x = k, k ∈ Z

f (x) = sin [nπx] is discontinuous when nπx = k, k ∈ Z

Thus, all the above functions have an infinite number of points of discontinuity. But, if (x) = |x| / x is discontinuous when x = 0 only.

### Differentiation

Differentiation is the other application of Differential Calculus and it is one of the most important concepts of the differential calculus that allows us to find a function that calculates the rate of change of one variable with respect to the other variable. The derivative of a function at a chosen input value describes the rate of change of the function near that input value. The process of finding the derivative of a function is called differentiation. Geometrically, the derivate at a particular point is the slope of the tangent line to the graph of the function at that point, provided that the derivative exists and is defined at that point.

Some of the differentiation formulae have been provided in the table below:

FunctionDifferential
xnnxn-1
k (Constant)0
exex
axaxlogx
logex1/x
Sinxcosx
Cosx-sinx
Tanxsec2x
Cotx-cosec2x
Secxsecxtanx
Cosecx-cosecxcotx

Example: If y is a function of x and log(x + y) = 2xy, then the value of y’(0) = ?

1. 1
2. -1
3. 2
4. 0

Solution: Given that log (x + y) = 2xy

Hence, at x = 0 we have log (y) = 0

This gives y = 1.

Now, to find  at (0, 1),

On differentiating the given equation with respect to x we have,

1/(x+y) . (1+ ) = 2x  + 2y.1

Hence,  = [2y(x + y) – 1]/[1-2(x + y)x]

So, ()|(0,1) = 1.

Example: If 2x + 2y = 2x+y then  has the value equal to

1. -2y/2x
2. 1/(1 – 2x)
3. 1 – 2y
4. 2x/2(1 – 2y)/(2-1)

Solution: The given function is 2x + 2y = 2x+y

Differentiating both sides we get

2x ln 2 + 2y ln 2 = 2x+y ln 2 (1 + )

Hence,  ( 2x+y - 2y) = 2- 2x+y

This gives = -2y/2x

A function can also be continuous when its graph is a single unbroken curve. This definition of a continuous function is useful when it is possible to draw the graph of a function so that just by the graph the continuity of the function can be judged.

Example: The number of values of x ∈ [0, 2] at which f (x) = ∣x − [1 / 2]∣ + |x − 1|+ tanx is not differentiable is

1. f(a) is defined.
2. exists (i.e., it is finite)

### Differential Equations

Differential equations is another most important application of Differential Calculus and carries 12 marks with approximately 4 to 6 questions from this topic in JEE Mains paper. A differential equation is an equation with a function and one or more of its derivatives. Example of a differential equation is

Example: Consider the differential equation

Statement 1: The substitution z = y2 transforms the above equation into a first order homogeneous differential equation.

Statement 2: The solution of this differential equation is:

(a) Statement 1 is false and statement 2 is true.

(b) Statement 1 is true and statement 2 is false.

(c) Both statements are true.

(d) Both statements are false.

Solution: (b) Put z = y2

### Tangent & Normal

Tangents: A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point.

Equation of tangent:

Normal: A normal to a curve is a line perpendicular to a tangent to the curve.

Equation of Normal:

### Short-Cut Methods and Tips to Study Differential Calculus

1. Memorize as many formulae as you can.
2. Use the method of hit and trial for the questions based on limits and continuity. This JEE Main 2020 trick will fetch you easy marks and is suitable for problems that sport complex orientation.
3. Learn the series expansions, most importantly of sinx, cosx, tanx, ex, logex etc.
4. Go through the detailed syllabus and divide each topic according to the time left to prepare for the examination.
5. Class Notes are the perfect example of an initiating step that helps you begin from scratch and build your concepts strong slowly. Pen down all the important information you receive in the class.
6. Candidates can purchase the online test series by some of the best and top coaching institutes in online mode only.
7. Online tutorials are good source of preparation as one can find unlimited video lectures on each subject. The content offered through these channels is reliable, given by field experts. Also, students can rely on these video lectures in case they’ve missed a topic or had doubts even after attending the lecture.
8. Candidates are advised to attempt at least 10- 15 previous year sample papers before appearing for the actual examination to understand the paper pattern, marking scheme and types of questions asked in the examination. This will give you an overview of the paper and the questions asked in the examination.

### JEE Main Previous Year Solved Questions on Differential Calculus

Question 1: The normal to the curve x2 + 2xy – 3y2 = 0 at (1,1): ( JEE Main 2013)

(a) meets the curve again in the third quadrant

(b) meets the curve again in the fourth quadrant

(c) does not meet the curve again

(d) meets the curve again in the second quadrant

Solution: (b) x2 + 3xy - xy - 3y2 = 0

x (x + 3y) –y (x + 3y) = 0

(x + 3y) (x – y) = 0

Normal at (1,1) will be x + y = 2.

Now, x + y = 2

x + 3y = 0

x = (3, -1) which is fourth quadrant.

Question 2: If (JEE Main 2012)

(a.) a = 1, b = 4.

(b.) a = 1, b = -4.

(c.) a = 2, b = -3.

(d.) a = 3, b = 3.

Solution: (b)

Dividing the numerator and denominator by x,

(1 – a) = 0 and (1 – b – a) = 4

From above equations,

a = 1 and b = -4.

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