To find the second derivative of the function \( f(x) = \frac{x^3 + 1}{3x} \), we will first need to simplify the function and then calculate the first and second derivatives.
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Simplifying the function: \[ f(x) = \frac{x^3}{3x} + \frac{1}{3x} = \frac{x^2}{3} + \frac{1}{3x} \]
We can rewrite this as: \[ f(x) = \frac{1}{3}x^2 + \frac{1}{3} x^{-1} \]
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Finding the first derivative \( f'(x) \):
Using the power rule, the derivative of \( x^n \) is \( nx^{n-1} \): \[ f'(x) = \frac{1}{3} \cdot 2x^{1} + \frac{1}{3} \cdot (-1)x^{-2} \] \[ f'(x) = \frac{2}{3} x - \frac{1}{3} x^{-2} \]
We can write \( f'(x) \) as: \[ f'(x) = \frac{2}{3} x - \frac{1}{3 x^2} \]
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Finding the second derivative \( f''(x) \):
We differentiate \( f'(x) \): \[ f''(x) = \frac{d}{dx}\left(\frac{2}{3} x\right) + \frac{d}{dx}\left(-\frac{1}{3 x^2}\right) \] \[ = \frac{2}{3} + \left(-\frac{1}{3} \cdot (-2)x^{-3}\right) = \frac{2}{3} + \frac{2}{3 x^3} \]
Therefore, we can write the second derivative as: \[ f''(x) = \frac{2}{3} + \frac{2}{3 x^3} \]
To summarize, the second derivative of the function \( f(x) = \frac{x^3 + 1}{3x} \) is: \[ f''(x) = \frac{2}{3} + \frac{2}{3 x^3} \]