Check sibling questions

Ex 6.4, 1 (xv) - Find approximate value of (32.15)^1/5 (upto 3 decimal

Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.4, 1 (xv) - Chapter 6 Class 12 Application of Derivatives - Part 4

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!


Transcript

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) 〖(32.15)〗^(1/5)Let 𝑦=(𝑥)^( 1/5) where 𝑥=32 & ∆𝑥=0. 15 Now, 𝑦=(𝑥)^( 1/5) Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^( 1/5) )/𝑑𝑥=1/5 𝑥^( 1/5 − 1) =1/5 𝑥^( (− 4)/( 5) )=1/(5〖 𝑥〗^( 4/5 ) ) Using ∆𝑦=𝑑𝑦/𝑑𝑥 ∆𝑥 ∆𝑦=1/(5 〖𝑥 〗^(4/5) ) ∆𝑥 Putting Values ∆𝑦=1/(5(32)^( 4/5) ) × (0. 15) ∆𝑦=1/(5(2)^( 5 × 4/5) ) × (0. 15) ∆𝑦=1/(5(2)^( 4) ) × (0. 15) ∆𝑦=1/(5 × 16) × 0. 15 ∆𝑦=(0. 15" " )/80 ∆𝑦=0. 00187 We know that ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) So, ∆𝑦=(𝑥+∆𝑥)^( 1/5)−〖𝑥 〗^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)−(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)−(2)^(5 × 1/5 ) 0. 00187=(32. 15)^( 1/5)−2 0. 00187+2=(32. 15)^( 1/5) 2. 00187=(32. 15)^( 1/5) (32. 15)^( 1/5)=2. 00187 Hence, Approximate Values of (32. 15)^( 1/5) is 𝟐. 𝟎𝟎𝟏𝟖𝟕 Hence, Approximate Values of (32. 15)^( 1/5) is 𝟐. 𝟎𝟎𝟏𝟖𝟕

Ask a doubt (live)
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.