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Home
JEE Maths
Differential Calculus Questions

Differential Calculus Questions 

Differential Calculus is a branch of mathematics that deals with the study of rates of change and slopes of curves. It primarily focuses on the concept of the derivative, which represents how a function changes as its input changes. From analyzing motion to optimizing real-world problems, differential calculus is essential in physics, engineering, economics, and more. It forms the foundation for advanced topics like integration, differential equations, and mathematical modeling.

1.0What is Differential Calculus?

Differential calculus studies how a function changes as its input changes. The central concept is the derivative, which represents the instantaneous rate of change or the slope of the tangent to a curve at a point.

Key Topics Covered in Differential Calculus

  • Derivative of a function
  • Rules of differentiation
  • Chain rule
  • Implicit differentiation
  • Higher-order derivatives
  • Tangents and normals
  • Increasing/decreasing functions
  • Maxima and minima

2.0Differential Calculus Formulas Recap

Concept

Formula

Power Rule

dxd​(xn)=nxn−1

Sum/Difference Rule

dxd​(f±g)=f′±g′

Product Rule

dxd​(uv)=u′v+uv′

Quotient Rule

dxd​(vu​)=v2u′v−uv′​

Chain Rule

dxdy​=dudy​⋅dxdu​

Derivative of sin x

cosx

Derivative of cos x

−sinx

Derivative of ex

ex

Derivative of ln(x)

x1​

3.0Differential Calculus Questions and Answers

Example 1: Find dxd​(3x4+2x2−5x+7)

Solution:
Using the power rule:

dxd​(3x4)=12x3,dxd​(2x2)=4x,dxd​(−5x)=−5,dxd​(7)=0

Final answer:12x3+4x−5


Example 2: Find dxd​(sin(x2))

Solution:
Let u=x2, then:

dxd​(sin(x2))=cos(x2)⋅dxd​(x2)=cos(x2)⋅2x=2xcos(x2)


Example 3: Find dxd​(xx2+1​)

Solution:

Use quotient rule: dxd​(vu​)=v2u′v−uv′​

Here, u=x2+1,v=x

u′=2x,v′=1

x2(2x)(x)−(x2+1)(1)​

=x22x2−x2−1​

=x2x2−1​ 


Example 4: Find the slope of the tangent to y=x3−3x at x = 1

Solution:

dxdy​=3x2−3

At x = 1,

dxdy​=3(1)2−3=0

So, the slope of the tangent = 0 (horizontal tangent).


Example 5: Find the maximum and minimum value of f(x)=x3−6x2+9x

Solution:

First derivative: f′(x)=3x2−12x+9

Set f'(x) = 0:

3x2−12x+9=0

⇒x2−4x+3=0

⇒x=1,3

Second derivative:

f''(x) = 6x - 12 

At x = 1, f''(1) = -6 ⇒ Maxima

At x = 3, f''(3) = 6 ⇒ Minima

4.0Challenging Differential Calculus Questions for Practice

Q1. If y=tan−1(1−x2​/x), find dxdy​

Q2. Find the derivative of xx using logarithmic differentiation.

Q3. Find the equation of the tangent to the curve y=ln(x2+1) at x = 1

Q4. Determine the points of inflection for y=x4−4x3+6x2

Q5. Find dxdy​ if x2+y2=25 using implicit differentiation.

Table of Contents


  • 1.0What is Differential Calculus?
  • 1.1Key Topics Covered in Differential Calculus
  • 2.0Differential Calculus Formulas Recap
  • 3.0Differential Calculus Questions and Answers
  • 4.0Challenging Differential Calculus Questions for Practice

Frequently Asked Questions

Differential calculus is a high-weightage topic in JEE Main and Advanced, often appearing in problems related to rate of change, optimization, and graphing.

While understanding is more important, knowing basic formulas and rules of differentiation helps solve problems faster and more accurately.

Finding slopes of curves, Tangents and Normals, Maximizing or minimizing functions, Analyzing growth, decay, motion, and optimization

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