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Chapter 6 Class 12 Application of Derivatives
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Ex 6.4, 1 (xiii) - Find approximate value upto 3 decimals - (81.5)^1/4

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

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiii) γ€–(81.5)γ€—^(1/4)Let 𝑦=(π‘₯)^( 1/4) where π‘₯=81 & βˆ†π‘₯=0. 5 Now, 𝑦=π‘₯^( 1/4) Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑(π‘₯^( 1/4) )/𝑑π‘₯=1/4 π‘₯^( 1/4 βˆ’ 1)=1/4 π‘₯^( (βˆ’ 3)/( 4)) =1/(4π‘₯^(3/4) ) βˆ† π‘₯ Using βˆ†π‘¦=𝑑𝑦/𝑑π‘₯ βˆ†π‘₯ βˆ†π‘¦=𝑑𝑦/(4π‘₯^(3/4) ) βˆ†π‘₯ Putting Values βˆ†π‘¦=1/(4(81)^( 3/4) ) Γ— (0. 5) βˆ†π‘¦=(0. 5)/(4 Γ— (3^( 4) )^( 3/4) ) βˆ†π‘¦=(0. 5)/(4 Γ— 3^( 3) ) βˆ†π‘¦=(0. 5)/(4 Γ— 27) βˆ†π‘¦=(0. 5)/108 βˆ†π‘¦=0. 0046 We know that βˆ†π‘¦=𝑓(π‘₯+βˆ†π‘₯)βˆ’π‘“(π‘₯) So, βˆ†π‘¦=(π‘₯+βˆ†π‘₯)^( 1/4)βˆ’π‘₯^( 1/4) Putting Values 0. 0046=(81+0. 5)^( 1/4)βˆ’(81)^( 1/4) 0. 0046=(81. 5)^( 1/4)βˆ’(3^4 )^( 1/4) 0. 0046=(81. 5)^( 1/4)βˆ’3 0. 0046+3=(81. 5)^( 1/4) 3. 0046=(81. 5)^( 1/4) Thus, Approximate Value of (81. 5)^( 1/4) is πŸ‘. πŸŽπŸŽπ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.