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. (viii) γ(255)γ^(1/4)Let π¦=(π₯)^(1/4) where π₯=256 & βπ₯=β1 Now, π¦=(π₯)^(1/4) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(γπ₯ γ^(1/4) )/ππ₯=1/4 π₯^( 1/4 β 1) =1/4 π₯^( (β3)/( 4))=1/(4 π₯^( 3/( 4)) ) Using βπ¦=ππ¦/ππ₯ βπ₯ βπ¦=1/(4γ π₯γ^( 3/( 4)) ) Γβπ₯ Putting Values βπ¦=1/(4 (256)^(3/4) ) Γ (β1) βπ¦=(β 1)/(4 (4^4 )^(3/4) ) βπ¦=(β 1)/(4 Γ 4^3 ) βπ¦=(β 1)/(4 Γ 64) βπ¦=(β 1)/256 βπ¦=β0. 0039 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) βπ¦=γ(π₯+βπ₯) γ^(1/4)βπ₯^( 1/4) Putting Values β0. 0039=γ(256+(β1)) γ^(1/4)β(256)^( 1/4) β0. 0039=γ(255) γ^(1/4)β(4^4 )^( 1/4) β0. 0039=γ(255) γ^(1/4)β4 β0. 0039=γ(255) γ^(1/4)β4 β0. 0039+4=γ(255) γ^(1/4) 3. 9961=γ(255) γ^(1/4) Thus, the Approximate Values of γ(255) γ^(1/4) is π. ππππ