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Let \(\rm f(x) = 2x^2+ \dfrac{1}{x}\) then f'(1) is
1. 1
2. 2
3. 3
4. 4

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Correct Answer - Option 3 : 3

Concept:

Formula used:

If f(x) = xn, then f'(x) = n xn - 1

Calculations:

Given:

 \(\rm f(x) = 2x^2+ \dfrac{1}{x}\)

Differentiating with respect to x, we get

⇒ f'(x) = 4x - \(\frac{1}{x^2}\)

Put x = 1

f'(x) = 4 × 1 - \(\frac{1}{1^2}\)

f'(-1) = 4 - 1 = 3

∴ The value of f'(1) is 3

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