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Saturday, 31 October

JEE Main · Maths · Continuity and Differentiability

Product Rule with Chain Rule

JEE Main asked Product Rule with Chain Rule once in the 108 shifts Lumi analysed, on 29 Jan 2024, Shift 1. It belongs to Chain Rule, in Continuity and Differentiability (Maths).

Updated 2026-10-24

Times asked
1
in 1 shift
Years
1 of 8
First asked
2024
the only year
Pattern
New since 2024
asked once so far

Questions on this idea a shift, by year

2017–2026. Asked in 2024. Years differ in how many shifts Lumi analysed, so each year is scaled to one shift of 25 Maths questions.

00.050.10’170’200’210’220’230.05’240’250’26

Raw count: 2017: 0 in 1 shift · 2020: 0 in 12 shifts · 2021: 0 in 24 shifts · 2022: 0 in 24 shifts · 2023: 0 in 12 shifts · 2024: 1 in 18 shifts · 2025: 0 in 8 shifts · 2026: 0 in 9 shifts.

Shift by shift

Each square is one analysed shift, in date order. Open a filled one for that paper.

  • 20260 of 9
  • 20250 of 8
  • 20241 of 18
  • 20230 of 12
  • 20220 of 24
  • 20210 of 24
  • 20200 of 12
  • 20170 of 1

Asked in that shift More than once Not asked

How JEE Main framed it

Single correct is the usual form. Skill tested: multi-step.

  • Derivative at a point of complicated product

Difficulty

Easy 0Medium 0Hard 1

The question was multiple choice.

Every question

The question

From the official JEE Main shift papers, newest first. Open one for the full question, the options and the official answer.

  1. 29 Jan 2024, Shift 1 · Q9Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3}. Then the value of f′(0)f'(0) is equal toHardSingle correct

Shift: 29 Jan 2024, Shift 1.

Same topic

Other ideas in Chain Rule

Chain Rule has 15 ideas JEE Main has asked; 8 of them came up in more than one question. The 12 most asked are below.

Topic analysisChapter analysisStudy this chapterAll repeated ideas