To find the approximate values using differentials, we can use the formula:
dy ≈ f'(x) * dx
1) √(25.01):
Let f(x) = √x
f'(x) = 1 / (2√x)
Let x = 25 and dx = 0.01.
dy ≈ f'(x) * dx
dy ≈ (1 / (2√25)) * 0.01
dy ≈ (1/10) * 0.01
dy ≈ 0.001
Approximate value: √(25.01) ≈ 5 + 0.001 = 5.001
2) 3√(125.1):
Let f(x) = 3√x
f'(x) = 1 / (3x^(2/3))
Let x = 125 and dx = 0.1.
dy ≈ f'(x) * dx
dy ≈ (1 / (3 * (125^(2/3)))) * 0.1
dy ≈ (1 / (3 * (125^(2/3)))) * 0.1
dy ≈ 0.004
Approximate value: 3√(125.1) ≈ 5 + 0.004 = 5.004
3) √(36.01):
Let f(x) = √x
f'(x) = 1 / (2√x)
Let x = 36 and dx = 0.01.
dy ≈ f'(x) * dx
dy ≈ (1 / (2√36)) * 0.01
dy ≈ (1/12) * 0.01
dy ≈ 0.000833
Approximate value: √(36.01) ≈ 6 + 0.000833 = 6.000833
Find the approximate values of the following using differentials
√(25.01)
3√(125.1)
√(36.01)
1 answer