

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Approximations (using Differentiation)
Question 1 (ii) Deleted for CBSE Board 2024 Exams You are here
Question 1 (iii) Deleted for CBSE Board 2024 Exams
Question 1 (iv) Deleted for CBSE Board 2024 Exams
Question 1 (v) Important Deleted for CBSE Board 2024 Exams
Question 1 (vi) Deleted for CBSE Board 2024 Exams
Question 1 (vii) Deleted for CBSE Board 2024 Exams
Question 1 (viii) Deleted for CBSE Board 2024 Exams
Question 1 (ix) Deleted for CBSE Board 2024 Exams
Question 1 (x) Deleted for CBSE Board 2024 Exams
Question 1 (xi) Important Deleted for CBSE Board 2024 Exams
Question 1 (xii) Deleted for CBSE Board 2024 Exams
Question 1 (xiii) Deleted for CBSE Board 2024 Exams
Question 1 (xiv) Important Deleted for CBSE Board 2024 Exams
Question 1 (xv) Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 9 (MCQ) Deleted for CBSE Board 2024 Exams
Approximations (using Differentiation)
Last updated at May 29, 2023 by Teachoo
Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ii) β49.5Let y = βπ₯ where x = 49 & β³ x = 0.5 Since y = βπ₯ ππ¦/ππ₯ = (π(βπ₯))/ππ₯ = 1/(2βπ₯) Now, βπ¦ = ππ¦/ππ₯ β³x = 1/(2βπ₯) (0.5) = 1/(2 Γ β49) Γ 0.5 = 1/(2 Γ 7) Γ 0.5 = 0.5/14 = 0.036 Also, βπ¦=π(π₯+βπ₯)βπ(π₯) Putting values βπ¦=β(π₯+βπ₯)ββπ₯ 0. 036=β(49+0. 5)ββ49 0. 036=β49.5β7 0. 036+7=β49.5 β49.5=7. 036 Hence, approximate value of β49.5 is 7.036