Answer
a. $y'=\frac{1}{x^2}$
b. See graph.
c. $f'(x)\gt0$ for all the x-values in the domain ($x\ne0$); $f'(x)$ will neither be zero or negative.
d. See explanations.
Work Step by Step
a. Given $y=f(x)=-\frac{1}{x}$, we have $y'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{-1/(x+h)+1/x}{h}=\lim_{h\to0}\frac{-x+(x+h)}{hx(x+h)}=\lim_{h\to0}\frac{1}{x(x+h)}=\frac{1}{x^2}$
b. See graph.
c. $f'(x)\gt0$ for all the x-values in the domain ($x\ne0$); $f'(x)$ will neither be zero or negative.
d. The function $f(x)$ increases when $x$ increases for all the x-values in the domain ($x\ne0$), and the corresponding derivative $f'(x)$ will be positive as shown in part-c. There are no region where the function $f(x)$ decreases when $x$ increases.
