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. (v) γ(0.999)γ^(1/10)Let π¦=γπ₯ γ^(1/10) where π₯=1 , βπ₯=β0. 001 Now, π¦=π₯^( 1/10) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(π₯^( 1/10) )/ππ₯=1/10 π₯^((β9)/10)=1/(10γ π₯γ^(9/10) ) Using βπ¦=ππ¦/ππ₯ βπ₯ Putting Values βπ¦= 1/(10γ π₯γ^(9/10) ) βπ₯ βπ¦= 1/(10 (1)^(9/10) ) Γ (β0. 001) βπ¦= 1/10 Γ(β0. 001) βπ¦=β0. 0001 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) βπ¦=γ(π₯+βπ₯) γ^(1/10)βπ₯^( 1/10) Putting Values βπ¦=(1+(β0. 001))^(1/10)β(1)^(1/10) β0. 0001=(0. 999)^(1/10)β1 β0. 0001+1=(0. 999)^(1/10) 0. 9999=(0. 999)^(1/10) Thus, the Approximate Value of (0. 999)^(1/10) is π. ππππ