Question
Download Solution PDFLet \(\rm f(x) = 2x^2+ \dfrac{1}{x}\) then f'(1) is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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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