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Last updated at April 19, 2021 by Teachoo

Example 21 Use differential to approximate β36.6Let y = βπ₯ where x = 36 & β³ x = 0.6 Since y = βπ ππ¦/ππ₯ = (π(βπ₯))/ππ₯ = 1/(2βπ₯) Now, βπ¦ = ππ¦/ππ₯ β³x = 1/(2βπ₯) (0.6) = 0.6/(2 Γ β36) = 0.6/(2 Γ 6) = 0.05 Now, β36.6=π¦+βπ¦ Putting values β36.6=β36+0.05 β36.6=6+0.05 β(ππ.π) = 6.05 Hence, approximate value of β36.6 is 6.05