How do you find the derivative of f(x)=(x-1)(x-2)(x-3)?

Aug 11, 2016

${f}^{'} \left(x\right) = \left(x - 3\right) \left(3 x - 5\right)$

Explanation:

$f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

$\text{let k(x),l(x),m(x) be equal (x-1),(x-2),(x-3);}$

$k \left(x\right) = \left(x - 1\right) \text{ ; } {k}^{'} \left(x\right) = 1$
$l \left(x\right) = \left(x - 2\right) \text{ ; } {l}^{'} \left(x\right) = 1$
$m \left(x\right) = \left(x - 3\right) \text{ ; } {m}^{'} \left(x\right) = 1$

${f}^{'} \left(x\right) = {k}^{'} \left(x\right) \cdot l \left(x\right) \cdot m \left(x\right) + {l}^{'} \left(x\right) \cdot k \left(x\right) \cdot m \left(x\right) + {m}^{'} \left(x\right) \cdot k \left(x\right) \cdot l \left(x\right)$

${f}^{'} \left(x\right) = 1 \cdot \left(x - 2\right) \cdot \left(x - 3\right) + 1 \cdot \left(x - 1\right) \cdot \left(x - 3\right) + 1 \cdot \left(x - 2\right) \cdot \left(x - 3\right)$

$\left(x - 2\right) \cdot \left(x - 3\right) = {x}^{2} - 3 x - 2 x + 6 = {x}^{2} - 5 x + 6$

$\left(x - 1\right) \cdot \left(x - 3\right) = {x}^{2} - 3 x - x + 3 = {x}^{2} - 4 x + 3$

${f}^{'} \left(x\right) = {x}^{2} - 5 x + 6 + {x}^{2} - 4 x + 3 + {x}^{2} - 5 x + 6$

${f}^{'} \left(x\right) = 3 {x}^{2} - 14 x + 15$

${f}^{'} \left(x\right) = \left(x - 3\right) \left(3 x - 5\right)$