By definition,
f ′ ( 2 ) = lim x → 2 f ( x ) − f ( 2 ) x − 2 = lim x → 2 1 1 − 2 x − 1 1 − 4 x − 2 = lim x → 2 1 1 − 2 x + 1 3 x − 2 = lim x → 2 ( 1 x − 2 3 + ( 1 − 2 x ) 3 ( 1 − 2 x ) ) = lim x → 2 ( 1 x − 2 − 2 ( x − 2 ) 3 ( 1 − 2 x ) ) = lim x → 2 − 2 3 ( 1 − 2 x ) = − 2 3 ⋅ ( − 3 ) = 2 9 . {\displaystyle {\begin{aligned}f'(2)&=\lim _{x\to 2}{\frac {f(x)-f(2)}{x-2}}\\&=\lim _{x\to 2}{\frac {{\frac {1}{1-2x}}-{\frac {1}{1-4}}}{x-2}}\\&=\lim _{x\to 2}{\frac {{\frac {1}{1-2x}}+{\frac {1}{3}}}{x-2}}\\&=\lim _{x\to 2}\left({\frac {1}{x-2}}{\frac {3+(1-2x)}{3(1-2x)}}\right)\\&=\lim _{x\to 2}\left({\frac {1}{x-2}}{\frac {-2(x-2)}{3(1-2x)}}\right)\\&=\lim _{x\to 2}{\frac {-2}{3(1-2x)}}\\&={\frac {-2}{3\cdot (-3)}}\\&={\frac {2}{9}}.\end{aligned}}}
Alternatively,
f ′ ( 2 ) = lim h → 0 f ( 2 + h ) − f ( 2 ) h = lim h → 0 1 1 − 4 − 2 h − 1 1 − 4 h = lim h → 0 1 − 3 − 2 h + 1 3 h = lim h → 0 ( 1 h 3 − 3 − 2 h 3 ( − 3 − 2 h ) ) = lim h → 0 ( 1 h − 2 h 3 ( − 3 − 2 h ) ) = lim h → 0 2 3 ( 3 + 2 h ) = 2 9 . {\displaystyle {\begin{aligned}f'(2)&=\lim _{h\to 0}{\frac {f(2+h)-f(2)}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {1}{1-4-2h}}-{\frac {1}{1-4}}}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {1}{-3-2h}}+{\frac {1}{3}}}{h}}\\&=\lim _{h\to 0}\left({\frac {1}{h}}{\frac {3-3-2h}{3(-3-2h)}}\right)\\&=\lim _{h\to 0}\left({\frac {1}{h}}{\frac {-2h}{3(-3-2h)}}\right)\\&=\lim _{h\to 0}{\frac {2}{3(3+2h)}}\\&={\frac {2}{9}}.\end{aligned}}}
