곱함수의 미분계수

[출제의도] 미분계수 계산하기 $\begin{aligned} f' \left( x \right) & = \left( x - 2 \right) \left( x - 3 \right) \cdots \left( x - 10 \right) \\ & + \left( x - 1 \right) \left( x - 3 \right) \cdots \left( x - 10 \right) \\ & \vdots \\ & + \left( x - 1 \right) \left( x - 2 \right) \cdots \left( x - 9 \right) \end{aligned}$ $f' \left( 1 \right) = \left( 1 - 2 \right) \left( 1 - 3 \right) \cdots \left( 1 - 10 \right)$ $f' \left( 4 \right) = \left( 4 - 1 \right) \left( 4 - 2 \right) \left( 4 - 3 \right) \left( 4 - 5 \right) \cdots \left( 4 - 10 \right)$ $\displaystyle \frac{f' \left( 1 \right)}{f' \left( 4 \right)} = \frac{( - 7 ) \times ( - 8 ) \times ( - 9 )}{3 \times 2 \times 1} = - 84$ (별해) $\displaystyle \begin{aligned} f' \left( 1 \right) & = \lim\limits _{x \rightarrow 1} {\frac{f ( x ) - f ( 1 )}{x - 1}} \end{aligned}$ $\displaystyle = \lim\limits _{x \rightarrow 1} {\left\{ ( x - 2 ) ( x - 3 ) ( x - 4 ) \cdots ( x - 10 ) \right\}}$ $= - 9 !$ $\displaystyle \begin{aligned} f' \left( 4 \right) & = \lim\limits _{x \rightarrow 4} {\frac{f ( x ) - f ( 4 )}{x - 4}} \end{aligned}$ $\displaystyle = \lim\limits _{x \rightarrow 4} {\left\{ ( x - 1 ) ( x - 2 ) ( x - 3 ) ( x - 5 ) \cdots ( x - 10 ) \right\}}$ $= 3 ! \times 6 !$ $\displaystyle \therefore \frac{f' \left( 1 \right)}{f' \left( 4 \right)} = - 84$