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Use the definition of the derivative (no differentiation rules) to find f ′ ( 2 ) {\displaystyle \displaystyle f'(2)} , where f ( x ) = 1 x {\displaystyle \displaystyle f(x)={\frac {1}{x}}} .
f ( x ) = 1 x {\displaystyle f(x)={\frac {1}{x}}}
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
f ′ ( x ) = lim h → 0 1 x + h − − 1 x h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {{\frac {1}{x+h}}-{\frac {-1}{x}}}{h}}}
f ′ ( x ) = lim h → 0 x x ( x + h ) − x + h x h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {{\frac {x}{x(x+h)}}-{\frac {x+h}{x}}}{h}}}
f ′ ( x ) = lim h → 0 − h x ( x + h ) h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {\frac {-h}{x(x+h)}}{h}}}
f ′ ( x ) = lim h → 0 − h x ( x + h ) h {\displaystyle f'(x)=\lim _{h\to 0}{\frac {-h}{x(x+h)h}}}
f ′ ( x ) = lim h → 0 − 1 x ( x + h ) {\displaystyle f'(x)=\lim _{h\to 0}{\frac {-1}{x(x+h)}}}
f ′ ( x ) = − 1 x ( x + ( 0 ) ) {\displaystyle f'(x)={\frac {-1}{x(x+(0))}}}
f ′ ( x ) = − 1 x 2 {\displaystyle f'(x)={\frac {-1}{x^{2}}}}
Plugging in 2 we get, -1/4
, where 
.
