JEE Main Study Notes for Differential Calculus: Differential Calculus is a branch of mathematics that deals with the rate of change of one quantity with respect to another. Say, In a particular direction with respect to time velocity is the rate of change of distance. If a function is f(x), then the differential equation is f′(x) = dy/dx. (where x≠0).

• It is the most scoring part in JEE Main. A minimum of 6-7 questions are asked on the topic of Differential Calculus.
• JEE Main Mathematics Question Paper has around 28 marks from Differential Calculus section.
• Some important topics from Differential Calculus are Differentiation, Limits, Continuity and Differentiability, Differential Equations, Tangent and Normal, Maxima and Minima, etc.

Quick Links:

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.

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.

Must Read:

• JEE Main Study Notes for Matrices & Determinants
• JEE Main Study Notes for Mathematical Reasoning

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.

Check:

• JEE Main Study Notes for Integral Calculus
• JEE Main Study Notes on Differential Calculus

Differentiation

Differentiation is the other application of Differential Calculus and it is one of the most important concepts of 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 derivative 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:

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/2y (1 – 2y)/(2x -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) = 2x - 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)

Must Read:

• JEE Main Study Notes for Trigonometry
• JEE Main Study Notes for Coordinate Geometry

Differential Equations

Differential equations are another most important application of Differential Calculus and carry 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. An 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: 

Read: JEE Main Probability Study Notes

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 from some of the best and top coaching institutes in online mode only. Check JEE Main 2022 Best Books for Preparation
7. Online tutorials are a 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. Check JEE Main 2022 Preparation Apps
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.

Must Read: Detailed JEE Main 2022 Mathematics Preparation Tips

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 the 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.

Quick Links: