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Application of Derivatives

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Short Answer Type

381.

Use differentials to approximate fourth root of $17 over 81$ .

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382.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 31.9 right parenthesis to the power of 1 fifth end exponent$

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383.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 0.999 right parenthesis to the power of 1 over 10 end exponent$

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384.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 32.15 right parenthesis to the power of 1 fifth end exponent$

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385.

Use differentials, find the approximate value of each of the following upto 3 places of decimal:
$left parenthesis 33 right parenthesis to the power of negative 1 fifth end exponent$

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386.

If y = x⁴ – 10 and if x changes from 2 to 1.99, what is the approximate change in y?

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387. If y = x⁴ + 10 and x changes from 2 to 1.99, find the approximate change in y.

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388.

Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2 .

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389.

Find the approximate value of f (5.001), where f (x) = x³ –7x² + 15.

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390.
Find the approximate value of f(3.02), where f (x) = 3x² + 5x + 3 .

f (x) = 3 x² + 5x + 3
Let x = 3 and ∆x = 0.02.
∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y = f (x) + f ' (x) ∆x (as dx = ∆x)
⇒ f (x + ∆x) = f (x) + (6 x + 5) ∆x
∴ f (3.2) = f(3) + {6(3) + 5} (0.02)
= [3 (3)² + 5(3) + 3] + (18 + 5) (0.02)
= (27 + 15 + 3)+ 23 (0.02)
= 45 + 0.46 = 45.046
∴ approximate value of f (3.02) is 45.46.

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