Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at April 15, 2021 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Transcript

Ex 6.4, 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and โ x = 0.01 Given f(x) = 4x2 + 5x + 2 fโ(x) = 8x + 5 Now, โ๐ฆ = fโ(x) โ x = (8x + 5) 0.01 Also, โy = f (x + โx) โ f(x) f(x + โ x) = f (x) + โ๐ฆ f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 ๐ (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21

Ex 6.4

Ex 6.4, 1 (i)
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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.