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. (xiii) γ(81.5)γ^(1/4)Let π¦=(π₯)^( 1/4) where π₯=81 & βπ₯=0. 5 Now, π¦=π₯^( 1/4) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(π₯^( 1/4) )/ππ₯=1/4 π₯^( 1/4 β 1)=1/4 π₯^( (β 3)/( 4)) =1/(4π₯^(3/4) ) β π₯ Using βπ¦=ππ¦/ππ₯ βπ₯ βπ¦=ππ¦/(4π₯^(3/4) ) βπ₯ Putting Values βπ¦=1/(4(81)^( 3/4) ) Γ (0. 5) βπ¦=(0. 5)/(4 Γ (3^( 4) )^( 3/4) ) βπ¦=(0. 5)/(4 Γ 3^( 3) ) βπ¦=(0. 5)/(4 Γ 27) βπ¦=(0. 5)/108 βπ¦=0. 0046 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) So, βπ¦=(π₯+βπ₯)^( 1/4)βπ₯^( 1/4) Putting Values 0. 0046=(81+0. 5)^( 1/4)β(81)^( 1/4) 0. 0046=(81. 5)^( 1/4)β(3^4 )^( 1/4) 0. 0046=(81. 5)^( 1/4)β3 0. 0046+3=(81. 5)^( 1/4) 3. 0046=(81. 5)^( 1/4) Thus, Approximate Value of (81. 5)^( 1/4) is π. ππππ