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. (vi) γ(15)γ^(1/4)Let π¦=γπ₯ γ^(1/4) where π₯=16 , βπ₯=β1 Now, π¦=π₯^( 1/4) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(π₯^( 1/4) )/ππ₯=1/4 π₯^( (1 β 4)/4 )=1/4 π₯^( (β3)/( 4) ) Using βπ¦=ππ¦/ππ₯ βπ₯ βπ¦=1/(4(π₯)^( 3/4) ) βπ₯ Putting Values βπ¦=1/(4(16)^( 3/4) ) . (β1) βπ¦=1/(4(2^4 )^( 3/4) ) (β1) βπ¦=(β1)/(4 Γ 2^3 ) βπ¦=(β1)/(4 Γ 8) βπ¦=(β1)/32 βπ¦=β0. 03125 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) βπ¦=(π₯+βπ₯)^(1/4)β(π₯)^(1/4) Putting Values β0. 03125=(16+(β1))^( 1/4)βγ(16) γ^(1/4) β0. 03125=(16β1)^( 1/4)β(2)^(4 Γ 1/4) β0. 03125=(15)^( 1/4)β2 β0. 03125+2=(15)^( 1/4) β0. 03125+2=(15)^( 1/4) 1. 96875=(15)^( 1/4) Thus, Approximate Value of (15)^( 1/4) is π. πππππ