


Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 6.4
Ex 6.4, 1 (ii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (iii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (iv) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (v) Important Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (vi) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (vii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (viii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (ix) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (x) Deleted for CBSE Board 2023 Exams You are here
Ex 6.4, 1 (xi) Important Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (xii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (xiii) Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (xiv) Important Deleted for CBSE Board 2023 Exams
Ex 6.4, 1 (xv) Deleted for CBSE Board 2023 Exams
Ex 6.4,2 Deleted for CBSE Board 2023 Exams
Ex 6.4,3 Important Deleted for CBSE Board 2023 Exams
Ex 6.4,4 Deleted for CBSE Board 2023 Exams
Ex 6.4,5 Important Deleted for CBSE Board 2023 Exams
Ex 6.4,6 Deleted for CBSE Board 2023 Exams
Ex 6.4,7 Deleted for CBSE Board 2023 Exams
Ex 6.4,8 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 6.4,9 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (x) 〖(401)〗^(1/2)Let 𝑦=𝑥^( 1/2) where 𝑥=400 & ∆𝑥=1 Now, 𝑑𝑦/𝑑𝑥=1/(2√𝑥) Using ∆𝑦=𝑑𝑦/𝑑𝑥 ∆𝑥 ∆𝑦=𝑑𝑦/(2√𝑥) ∆𝑥 Putting Values ∆𝑦=1/(2√400) ×(1) ∆𝑦=1/(2√400) ×(1) ∆𝑦=1/(2√(20^2 )) ∆𝑦=1/(2 × 20) ∆𝑦=1/40 ∆𝑦=0. 025 We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦= (𝑥+∆𝑥)^( 1/2)−(𝑥)^( 1/2) Putting Values 0. 025=(400+1)^( 1/(2 ))−〖(400) 〗^(1/2) 0. 025=(401)^( 1/2)−(20)^( 2 × 1/2) 0. 025=(401)^( 1/2)−20 0. 025+20=(401)^( 1/2) 20. 025=(401)^( 1/2) We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦= (𝑥+∆𝑥)^( 1/2)−(𝑥)^( 1/2) Putting Values 0. 025=(400+1)^( 1/(2 ))−〖(400) 〗^(1/2) 0. 025=(401)^( 1/2)−(20)^( 2 × 1/2) 0. 025=(401)^( 1/2)−20 0. 025+20=(401)^( 1/2) 20. 025=(401)^( 1/2)