Using differentials, find the the approximate value.
1. (29) to the power of 1/3
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The answer is explained in the attachment.
Hope this helps!
Hope this helps!
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Anonymous:
Awesome answer bhai.
Answered by
3
Hi,
Here is your answer,
(29)¹/³
= ( 27 + 2)¹/³ ∴ x = 27 , δx = 2
Let y = x¹/³
dy = 1/3 × x⁻²/³ × dx
dy = 1/3 × (27)⁻²/³ × (2)
= 1/3 × (3³)⁻²/³ × (2) { On cancelling 3 }
= 1/3 × 1/(3)² × (2)
= 2/(3)³
= 2/27
Now, by using 1st Principle
f(x + δx/Δx) = f(x) + f'(x) × δx/Δx
= (27)¹/³ + 2/27
= (3³)¹/³ + 2/27 { On cancelling 3 }
= 3 + 2/27
= 81 + 2/27
= 83/27
= 3.07
Hope it helps you !
Here is your answer,
(29)¹/³
= ( 27 + 2)¹/³ ∴ x = 27 , δx = 2
Let y = x¹/³
dy = 1/3 × x⁻²/³ × dx
dy = 1/3 × (27)⁻²/³ × (2)
= 1/3 × (3³)⁻²/³ × (2) { On cancelling 3 }
= 1/3 × 1/(3)² × (2)
= 2/(3)³
= 2/27
Now, by using 1st Principle
f(x + δx/Δx) = f(x) + f'(x) × δx/Δx
= (27)¹/³ + 2/27
= (3³)¹/³ + 2/27 { On cancelling 3 }
= 3 + 2/27
= 81 + 2/27
= 83/27
= 3.07
Hope it helps you !
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