Answer
a) $ f′(x)=\dfrac{-1}{x^2}$
b) $\dfrac{-1}{4}, \dfrac{-1}{4}$
Work Step by Step
a) Now, $ f′(x)=\lim_\limits{ h\to 0} \dfrac{f(x+h)-f(x)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{\dfrac{1}{x+h}-1/x}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{-1}{x(x+h)}$
or, $=6x^2+6x(0)+2(0)^2-1$
or, $ f′(x)=\dfrac{-1}{x^2}$
b) $ f′(-2)=\dfrac{-1}{(-2)^2}=\dfrac{-1}{4}$
and $ f′(2)=\dfrac{-1}{(2)^2}=\dfrac{-1}{4}$
