Example 22 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 22 Use differential to approximate (25)13 Let 𝑦 =𝑥13 Let 𝑥=27 and △𝑥=−2 Since 𝑦 =𝑥13 𝑑𝑥𝑑𝑦=𝑑𝑥13𝑑𝑥 𝑑𝑥𝑑𝑦=13 𝑥13 − 1 𝑑𝑥𝑑𝑦=13 𝑥−23 𝑑𝑥𝑑𝑦=13𝑥23 Now, ∆𝑦=𝑑𝑦𝑑𝑥∆𝑥 ∆𝑦= 13𝑥23∆𝑥 Putting values ∆𝑦= 132723−2 ∆𝑦= −23 ×3323 ∆𝑦= −23 × 32 ∆𝑦= −227 ∆𝑦=−0.074 Also, ∆𝑦=𝑓𝑥+∆𝑥 −𝑓𝑥 ∆𝑦=𝑥+∆𝑥13−𝑥13 Putting values −0.074=27−213−2713 −0.074=2513−33 × 13 −0.074=2513−3 2513=−0.074+3 2513=2.926 Hence, approximate value of 2513 is 𝟐.𝟗𝟐𝟔
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Chapter 6 Class 12 Application of Derivatives
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.