Approximations (using Differentiation)

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xi) γ(0.0037)γ^(1/2)Let y = βπ₯ Let x = 0.0036 & β³ x = 0.0001 Since y = βπ₯ ππ¦/ππ₯ = (π(βπ₯))/ππ₯ = 1/(2βπ₯) Now, βπ¦ = ππ¦/ππ₯ β³x = 1/(2βπ₯) (0.0001) = 1/(2β0.0036) (0.0001) = 1/(2 Γ 0.06) Γ (0.0001) = 0.0001/0.12 = 1/1200 = 0.000833 Also, βπ¦=π(π₯+βπ₯)βπ(π₯) So, βπ¦=(π₯+βπ₯)^( 1/2)β(π₯)^( 1/2) Putting Values 0. 000833=(0. 0036+0. 0001)^( 1/2)β(0. 0036)^( 1/2) 0. 000833=(0. 0037)^( 1/2)β(36/10000)^(1/2) 0. 000833=(0. 0037)^( 1/2)β(6/100)^( 2 Γ 1/2) 0. 000833=(0. 0037)^( 1/2)β(0. 06)^(2 Γ 1/2 ) 0. 000833=(0. 0037)^( 1/2)β(0. 06) 0. 000833+0. 06=(0. 0037)^( 1/2) 0. 060833=(0. 0037)^( 1/2) (0. 0037)^( 1/2)=0. 060833 Thus, the Approximate Value of (0. 0037)^( 1/2)=π. ππππππ