#### Answer

$$y'=\frac{1}{x^2}$$
The graphs of $y=f(x)$ (red) and $y=f'(x)$ (green) are plotted below.

#### Work Step by Step

$$y=-\frac{1}{x}$$
a) Find the derivative:
$$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{-\frac{1}{z}-(-\frac{1}{x})}{z-x}=\lim_{z\to x}\frac{\frac{1}{x}-\frac{1}{z}}{z-x}$$ $$y'=\lim_{z\to x}\frac{\frac{z-x}{xz}}{z-x}=\lim_{z\to x}\frac{z-x}{zx(z-x)}=\lim_{z\to x}\frac{1}{zx}$$ $$y'=\frac{1}{x\times x}=\frac{1}{x^2}$$
b) The graphs are plotted below. On the left side, the red line represents the graph of $y=f(x)$. On the right side, the green line represents the graph of $y=f'(x)$.
c) From the right graph, we see that on $(-\infty,\infty)$, $f'(x)$ is positive. There is no point or interval where $f'(x)$ equals $0$ or is negative.
d) From the left graph, we see that $f(x)$ increases as $x$ increases on $(-\infty,\infty)$. Only as $x$ passes $0$ does $f(x)$ change from $\infty$ to $-\infty$. This corresponds to the interval where $f'$ is positive.