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JEE Exam  >  JEE Test  >  Mathematics (Maths) Main & Advanced  >  Test: Derivatives: Product And Quotient Rule - JEE MCQ

Derivatives: Product And Quotient Rule - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Derivatives: Product And Quotient Rule (10 Questions)

You can prepare effectively for JEE Mathematics (Maths) for JEE Main & Advanced with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Derivatives: Product And Quotient Rule". These 10 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Test: Derivatives: Product And Quotient Rule - Question 1
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The derivate of the function 

Detailed Solution: Question 1

y = (1 + 1/x)/(1 - 1/x)
y = (x+1)/(x-1)
dy/dx = [(x-1)-(x+1)]/(x-1)2
= -2/(x-1)2

Test: Derivatives: Product And Quotient Rule - Question 2
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The derivate of the function 

Detailed Solution: Question 2

Given, y = (sinx+cosx)/(sinx−cosx)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)2 
[by quotient rule]
= [(−sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
− [(sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
= − [(sin2x + cos2x − 2sinxcosx + sin2x + cos2x + 2sinxcosx)]/(sinx−cosx)2
= −2/(sinx−cosx)2
= -2/(sin2x + cos2x - 2sinxcosx)
= - 2/(1 - sin2x)

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Test: Derivatives: Product And Quotient Rule - Question 3
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Find the derivative of the given equation f(x) = x3 + x2 + 3

Detailed Solution: Question 3

using these two rulesdifferentiation of  f(x) = x3 + x2 + 3 will be 3x2 + 2x

Test: Derivatives: Product And Quotient Rule - Question 4
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The derivate of the function 

Detailed Solution: Question 4

Test: Derivatives: Product And Quotient Rule - Question 5
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Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are

Detailed Solution: Question 5


Test: Derivatives: Product And Quotient Rule - Question 6
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The derivative of the function sec2 x is:

Test: Derivatives: Product And Quotient Rule - Question 7
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Find w'(z) = 

Detailed Solution: Question 7

w’(z) = [(2-z)dy/dz(3z+9)- (3z+9)dy/dz(2-x)]/(2-z)2
[(2-z)(3) - (3z+9)(-1)]/(2-z)2
(6-3z+3z+9)/(2-z)2
= 15/(2-z)2

Test: Derivatives: Product And Quotient Rule - Question 8
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Detailed Solution: Question 8

Test: Derivatives: Product And Quotient Rule - Question 9
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The derivate of  f(x) = 

Test: Derivatives: Product And Quotient Rule - Question 10
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The derivative of the function y = (px2+qx+r)/(ax+b) w.r.t. x is :

Detailed Solution: Question 10

y = (px2+qx+r)/(ax+b)
⇒ dy/dx = d/dx(px2+qx+r)/(ax+b)
= [(ax+b) d/dx(px2+qx+r) − (px2+qx+r) d/dx(ax+b)]/(ax+b)2
= [(ax+b)(2px+q) − (px2+qx+r).a]/(ax+b)2
= [(2apx2 + aqx + 2bpx + bq) − (apx2 + aqx + ar)]/(ax+b)2
= [apx2 + 2bpx + bq − ar]/(ax+b)2

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About this JEE Practice Test

This test contains 10 MCQs on Derivatives: Product And Quotient Rule with detailed solutions for each question. It is part of the Mathematics (Maths) for JEE Main & Advanced course on EduRev, designed to align with the current JEE syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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