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Use the definition of derivative to find $f^{\prime}(2)$ for $f(x)=x+\frac{1}{x}$.
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Using a reciprocal rule $f(x) = x - x^{-1}$ then differentiating this we get $f'(x) = 1 - x^{-2}$. Back to fractional notation we have $f'(x) = 1 - \frac{1}{x^2}$.

When $x=2$ the derivative $f'(2)$ which is the final answer is $1-\frac{1}{2^2}= 0.75$
by Wooden (194 points)
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