Answer
(a) $f'(x) = 1-\frac{1}{x^2}$
(b) When we compare the graphs of $f(x) = x+\frac{1}{x}$ and $f'(x) = 1-\frac{1}{x^2}$, the answer to part (a) seems reasonable.
Work Step by Step
(a) $f(x) = x + \frac{1}{x}$
We can find $f'(x)$:
$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{(x+h)+\frac{1}{x+h}-(x+\frac{1}{x})}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{h+\frac{1}{x+h}-\frac{1}{x}}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{\frac{hx(x+h)+x-(x+h)}{(x)(x+h)}}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{hx(x+h)-h}{(h)(x)(x+h)}$
$f'(x) = \lim\limits_{h \to 0} \frac{x(x+h)-1}{(x)(x+h)}$
$f'(x) = \frac{x(x+0)-1}{(x)(x+0)}$
$f'(x) = \frac{x^2-1}{x^2}$
$f'(x) = 1-\frac{1}{x^2}$
(b) When we compare the graphs of $f(x) = x+\frac{1}{x}$ and $f'(x) = 1-\frac{1}{x^2}$, the answer to part (a) seems reasonable.
