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Chapter 6 Class 12 Application of Derivatives
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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

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) γ€–(32.15)γ€—^(1/5)Let 𝑦=(π‘₯)^( 1/5) where π‘₯=32 & βˆ†π‘₯=0. 15 Now, 𝑦=(π‘₯)^( 1/5) Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(π‘₯^( 1/5) )/𝑑π‘₯=1/5 π‘₯^( 1/5 βˆ’ 1) =1/5 π‘₯^( (βˆ’ 4)/( 5) )=1/(5γ€– π‘₯γ€—^( 4/5 ) ) Using βˆ†π‘¦=𝑑𝑦/𝑑π‘₯ βˆ†π‘₯ βˆ†π‘¦=1/(5 γ€–π‘₯ γ€—^(4/5) ) βˆ†π‘₯ Putting Values βˆ†π‘¦=1/(5(32)^( 4/5) ) Γ— (0. 15) βˆ†π‘¦=1/(5(2)^( 5 Γ— 4/5) ) Γ— (0. 15) βˆ†π‘¦=1/(5(2)^( 4) ) Γ— (0. 15) βˆ†π‘¦=1/(5 Γ— 16) Γ— 0. 15 βˆ†π‘¦=(0. 15" " )/80 βˆ†π‘¦=0. 00187 We know that βˆ†π‘¦=𝑓(π‘₯+βˆ†π‘₯)βˆ’π‘“(π‘₯) So, βˆ†π‘¦=(π‘₯+βˆ†π‘₯)^( 1/5)βˆ’γ€–π‘₯ γ€—^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)βˆ’(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)βˆ’(2)^(5 Γ— 1/5 ) 0. 00187=(32. 15)^( 1/5)βˆ’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. πŸŽπŸŽπŸπŸ–πŸ• Hence, Approximate Values of (32. 15)^( 1/5) is 𝟐. πŸŽπŸŽπŸπŸ–πŸ•

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.