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. (xv) γ(32.15)γ^(1/5)Let π¦=(π₯)^( 1/5) where π₯=32 & βπ₯=0. 15 Now, π¦=(π₯)^( 1/5) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(π₯^( 1/5) )/ππ₯=1/5 π₯^( 1/5 β 1) =1/5 π₯^( (β 4)/( 5) )=1/(5γ π₯γ^( 4/5 ) ) Using βπ¦=ππ¦/ππ₯ βπ₯ βπ¦=1/(5 γπ₯ γ^(4/5) ) βπ₯ Putting Values βπ¦=1/(5(32)^( 4/5) ) Γ (0. 15) βπ¦=1/(5(2)^( 5 Γ 4/5) ) Γ (0. 15) βπ¦=1/(5(2)^( 4) ) Γ (0. 15) βπ¦=1/(5 Γ 16) Γ 0. 15 βπ¦=(0. 15" " )/80 βπ¦=0. 00187 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) So, βπ¦=(π₯+βπ₯)^( 1/5)βγπ₯ γ^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)β(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)β(2)^(5 Γ 1/5 ) 0. 00187=(32. 15)^( 1/5)β2 0. 00187+2=(32. 15)^( 1/5) 2. 00187=(32. 15)^( 1/5) (32. 15)^( 1/5)=2. 00187 Hence, Approximate Values of (32. 15)^( 1/5) is π. πππππ Hence, Approximate Values of (32. 15)^( 1/5) is π. πππππ